M.C.Q (1 Marks)

MCQ 11 Mark

He area of the region bounded by the parabola $y = x^2$ and $y = |x|$ is:

A
$3$
B
$\frac{1}{2}$
✓
$\frac{1}{3}$
D
$2$

Answer

Correct option: C.

$\frac{1}{3}$

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MCQ 21 Mark

The area of the region bounded by the curve $x = y^2 - 2$ and $x = y$ is:

A
$\frac { 9 }{ 4 }$
B
$9$
✓
$\frac { 9 }{ 2 }$
D
$\frac { 9 }{ 7 }$

Answer

Correct option: C.

$\frac { 9 }{ 2 }$

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MCQ 31 Mark

The area of the ellipse $\frac{\text{x}2}{9}+\frac{\text{y}^2}{4}=1$ in first quadrant is $6\pi\ \text{sq. units}.$The ellipse is rotated about its centre in anti$-$clockwise direction till its major axis coincides with $y-$axis. Now the area of the ellipse in first Quadrant is $\pi\ \text{sq. units}.$

A
$2$
✓
$4$
C
$6$
D
$8$

Answer

Correct option: B.

$4$

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MCQ 41 Mark

Area bounded by the lines $y = |x| - 2$ and $y = 1 - |x - 1|$ is equal to:

✓
$4 \ sq.$ units
B
$6 \ sq.$ units
C
$2 \ sq.$ units
D
$8 \ sq.$ units

Answer

Correct option: A.

$4 \ sq.$ units

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MCQ 51 Mark

Area of the region bounded by $y = |x – 1|$ and $y = 1$ is:

A
$2\text{ sq.}\text{ units}$
✓
$1\text{ sq.}\text{ units}$
C
$\frac{1}{2}\text{ sq.}\text{ units}$
D
None of these

Answer

Correct option: B.

$1\text{ sq.}\text{ units}$

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MCQ 61 Mark

The area of the region $\{(\text{x},\text{y}):\text{x}^2+\text{y}^2\leq1\leq\text{x}+\text{y}\}$ is:

A
$\frac{\pi}{5}$
B
$\frac{\pi}{4}$
✓
$\frac{\pi}{2}-\frac{1}{2}$
D
$\frac{\pi^2}{2}$

Answer

Correct option: C.

$\frac{\pi}{2}-\frac{1}{2}$

None of the given option is correct.
To find the points of intersection of the line
and the circle substitute $y = 1 - x$ in $x^2 + y^2 = 1$,
we get $A(0, 1)$ and $B(1, 0).$
Therefore, the required area of the shaded region,
$\text{A} = \int\limits^1_0(\text{y}_1-\text{y}_2)\text{dx}$ $\big($Where, $\text{y}_1=\sqrt{1-\text{x}^2}$ and $\text{y}_2=1-\text{x}\big)$
$= \int\limits^1_0\Big[\big(\sqrt{1-\text{x}^{2}}\big)-(1-\text{x})\Big]\text{dx}$
$=\int\limits^1_0\Big(\sqrt{1-\text{x}^2}-1+\text{x}\Big)\text{dx}$
$=\Big[\frac{\text{x}}{2}\sqrt{1-\text{x}^2}+\frac{1}{2}\sin^{-1}(\text{x})-\text{x}+\frac{\text{x}^2}{2}\Big]^1_0$
$=\bigg[\frac{1}{2}\sqrt{1-1^2}+\frac{1}{2}\sin^{-1}(1)-(1)+\frac{(1)^2}{2}\bigg]\\-\bigg[\frac{(0)}{2}\sqrt{1-(0)^2}+\frac{1}{2}\sin^{-1}(0)-(0)+\frac{(0)^2}{2}\bigg]$
$=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)$ square units

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MCQ 71 Mark

Choose the correct answer : Smaller area enclosed by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ is:

A
$2(\pi-2)$
✓
$\pi-2$
C
$2\pi-1$
D
$2(\pi+2).$

Answer

Correct option: B.

$\pi-2$

Step $I$. Equation of circle is $x^2 + y^2 = 2^2 ...(i)$

$\Rightarrow\text{y}=\sqrt{2^2-\text{x}^2}\dots(\text{ii})$ Also, equation of the line is $x + y = 2 ...(iii)$ Table of values

$x$	$0$	$2$
$y$	$2$	$0$

Therefore graph of equation $(iii)$ is the straight line joining the points $(0, 2)$ and $(2, 0)$.
Step $II$. From the graph of circle $(i)$ and straight line $(iii),$ it is clear that points of intersections of circle $(i)$ and straight line $(iii)$ are $A(2, 0)$ and $B(0, 2).$
Step $III$. Area $\text{OACB},$ bounded by circle $(i)$ and coordinate axes in first quadrant
$=\Bigg|\int\limits^2_0\text{y dx}\Bigg|=\Bigg|\int\limits^2_0\sqrt{2^2-\text{x}^2}\text{ dx}\Bigg|$
$=\Big(\frac{\text{x}}{2}\sqrt{2^2-\text{x}^2}+\frac{2^2}{2}\sin^{-1}\frac{\text{x}}{2}\Big)^2_0$
$=\Big(\frac22\sqrt{4-4}+2\sin^{-1}1\Big)-\Big(0+2\sin^{-1}0\Big)$
$=0+2\Big(\frac{\pi}{2}\Big)-2(0)=\pi\text{ sq. units}\dots(\text{iv})$
Step $IV$. Area of triangle $\text{OAB},$ bounded by straight line $(iii)$ and coordinate axes
$=\Bigg|\int\limits^2_0\text{y dx}\Bigg|=\Bigg|\int\limits^2_0(2-\text{x})\text{ dx}\Bigg|$
$=\Big(2\text{x}-\frac{\text{x}^2}{2}\Big)^2_0$
$=(4-2)-(0-0)=2\text{ sq. units}\dots(\text{v})$
Step $V$. Required shaded area $=$ Area $\text{OACB}$ given by $(iv) -$ Area of triangle $\text{OAB}$ by $(v)$
$=(\pi-2)\text{ sq. units}$
Therefore, option $(B)$ is correct.

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MCQ 81 Mark

The area of the region bounded by the curve $\text{y}=\sqrt{16-\text{x}^2}$ and $x-$axis is:

✓
$8\pi\text{ sq.}\text{units}$
B
$20\pi\text{ sq.}\text{units}$
C
$16\pi\text{ sq.}\text{units}$
D
$256\pi\text{ sq.}\text{units}$

Answer

Correct option: A.

$8\pi\text{ sq.}\text{units}$

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MCQ 91 Mark

The area of the triangle formed by the tangent and normal at the point $(1,\sqrt{3})$ on the circle $x^2 + y^2 = 4$ and the $x-$ axis is:

A
$3\text{ sq.}\text{ units}$
✓
$2\sqrt{3}\text{ sq.}\text{ units}$
C
$3\sqrt{2}\text{ sq.}\text{ units}$
D
$4\text{ sq.}\text{ units}$

Answer

Correct option: B.

$2\sqrt{3}\text{ sq.}\text{ units}$

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MCQ 101 Mark

The area bounded by the curve $2x^2 + y^2 = 2$ is :

A
$\pi\text{ sq}.\text{units}$
✓
$\sqrt{2}\pi\text{ sq}.\text{units}$
C
$\frac{\pi}{2}\text{sq}.\text{units}$
D
$2\pi\text{ sq}.\text{units}$

Answer

Correct option: B.

$\sqrt{2}\pi\text{ sq}.\text{units}$

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MCQ 111 Mark

If $\text{y}=2\sin\text{x}+\sin2\text{x}$ for $0≤\text{x}≤2\pi,$ then the area enclosed by the curve and $x-$axis is:

A
$\frac{9}{2}\text{sq.}\text{units}$
B
$\text{8 sq. units}$
✓
$\text{12 sq. units}$
D
$\text{4 sq. units}$

Answer

Correct option: C.

$\text{12 sq. units}$

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MCQ 121 Mark

Choose the correct answer in the following. Area bounded by the curve $y = x^3,$ the $x-$ axis and the ordinates $x = –2$ and $x = 1$ is :

A
$-9$
✓
$-\frac{15}{4}$
C
$\frac{15}{4}$
D
$\frac{17}{4}.$

Answer

Correct option: B.

$-\frac{15}{4}$

$\text{Required area}=\int\limits^1_{-2}\text{y dx}$
$=\int\limits^1_{-2}\text{x}^3\text{dx}$
$=\Big[\frac{\text{x}^4}{4}\Big]^1_{-2}$
$=\Big[\frac14-\frac{(-2)^4}{4}\Big]$
$=\Big(\frac14-4\Big)=-\frac{15}{4}\text{ units}$
Thus, the correct answer is $B.$

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MCQ 131 Mark

The area bounded by the line $y = 2x - 2, y = -x$ and $x-$axis is given by:

A
$\frac{9}{2}\text{sq}.\text{units}$
B
$\frac{43}{6}\text{sq}.\text{units}$
C
$\frac{35}{6}\text{ sq}.\text{units}$
✓
None of these

Answer

Correct option: D.

None of these

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MCQ 141 Mark

The area bounded by the curve $\text{y}=\log_{\text{e}}\text{x}$ and x-axis and the straight line x = e is:

A
$\text{e sq. units}$
✓
$1\text{ sq. units}$
C
$1-\frac{1}{\text{e}}\text{ sq. units}$
D
$1+\frac{1}{\text{e}}\text{ sq. units}$

Answer

Correct option: B.

$1\text{ sq. units}$

The point of intersection of the curve and the straight line is A(e, 1). Therefore, the area of the required region ABC,

$\text{A} = \int\limits^1_0(\text{x}_1-\text{x}_2)\text{dy}$ $(\text{where}, \text{x}_1 = \text{e}\text { and }\text{x}_2 = \text{e}_{\text{y}})$

$= \int\limits^1_0(\text{e}-\text{e}^{\text{y}})\text{dy}$

$=\big [\text{ey}-\text{e}^{\text{y}}\big]^1_0$

$=\big\{\text{e}(1)-\text{e}^{(1)}\big\} -\big \{\text{e}(0)-\text{e}^{(0)}\big\}$

$= \text{e}-\text{e}+1$

$= 1 \text{ square unit}$

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MCQ 151 Mark

Area bounded by the curve $\text{y}=\cos\text{x}$ between $\text{x}=0$ and $\text{x}=3\frac{\pi}{2}$ is:

A
$1 \ sq.$ unit
B
$2 \ sq.$ units
✓
$3 \ sq.$ units
D
$4 \ sq.$ units

Answer

Correct option: C.

$3 \ sq.$ units

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MCQ 161 Mark

The area included between the parabolas $y^2 = 4x$ and $x^2 = 4y$ is:

A
$\frac{8}{3}\text{sq}\text{ unit}$
B
$8\text{sq}\text{ unit}$
✓
$\frac{16}{3}\text{sq}\text{ unit}$
D
$12\text{sq}\text{ unit}$

Answer

Correct option: C.

$\frac{16}{3}\text{sq}\text{ unit}$

We know that, the area of region bounded by the parabolas $y^2 = 4ax$ and $= 4by$ is
$=\frac{16}{3}\text{ab}\text{ sq.}\text{ unit.}$
Therefore, $y^2 = 4ax$ and $x^2 = 4y$ is
$=\frac{16}{3}\text{ sq.}\text{ unit.}$
$(\because\text{a}=1,\text{b}=1)$

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MCQ 171 Mark

The area of the region bounded by the ellipse $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{16}=1$ is:

✓
$20\pi\text{ sq}.\text{units}$
B
$20^2\pi\text{ sq}.\text{units}$
C
$16^2\pi\text{ sq}.\text{units}$
D
$25\pi\text{ sq}.\text{units}$

Answer

Correct option: A.

$20\pi\text{ sq}.\text{units}$

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MCQ 181 Mark

Area bounded by the curve $y = x^3,$ the $x-$ axis and the ordinates $x = -2$ and $x = 1$ is:

A
$-9$
B
$\frac{-15}{4}$
C
$\frac{15}{4}$
✓
$\frac{17}{4}$

Answer

Correct option: D.

$\frac{17}{4}$

$x = -2$ and $x = 1$ intersect the curve $y = x^3$ at $A(-2, -8)$ and $B(1, 1)$ respectively
If $P(x, y_1)$ lies on $OA\ \ O(x, y_2)$ lies on curve $OB$
Then, $y_1 > 0 $
$\Rightarrow |y_1| = y_1$
$\ \ y_2 < 0 $
$\Rightarrow |y_2| = -y_2$
Area of curve bound by the two lines $=$ shaded are $\text{(OADO)}\ +$ shaded area $\text{(OCBO)}$
$= \int\limits^0_{-2}|\text{y}_2|\text{dx}+\int\limits^1_0|\text{y}_1|\text{dx}$
$= \int\limits^0_{-2}-\text{y}_2\text{dx}+\int\limits^1_0\text{y}_1\text{dx}$
$=\int\limits^0_{-2}-(\text{x}^3)\text{dx}+\int\limits^1_0\text{x}^3\text{dx}$
$= \Big[-\frac{\text{x}^4}{4}\Big]^0_{-2}+\Big[\frac{\text{x}^4}{4}\Big]^1_0$
$= 0-\Big(-\frac{16}{4}\Big)+\Big(\frac{1}{4}-0\Big)$
$= 4+\frac{1}{4}$
$=\frac{17}{4}\text{ sq. units}$

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MCQ 191 Mark

Find the area of the region bounded by the curves $y = x^3,$ the line $x = 2, x = 5$ and the $x -$ axis?

A
$173.50$
B
$230.25$
C
$175.35$
✓
$152.25$

Answer

Correct option: D.

$152.25$

$\int\limits\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n-1}}{\text{n+1}}+\text{c}$
Here, we have to find the area of the region bounded by the curves $y = x^3$, the line $x = 2, x = 5$ and the $x -$ axis
So, the area enclosed by the given curves is given by $\int\limits^3_2\text{x}^3\text{dx}$
As we know that, $\int\limits\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n}}{\text{n+1}}+\text{c}$
$\Rightarrow\int\limits^5_2\text{x}^3\text{dx}=\Big[\frac{\text{x}^4}{4}\Big]^5_4$
$\Rightarrow\int\limits^5_2\text{x}^3\text{dx}=\frac{1}{4}(625-16)$
$=152.25$

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MCQ 201 Mark

Area bounded by parabola $y^2 = x$ and straight line $2y = x$ is:

✓
$43$
B
$1$
C
$23$
D
$13$

Answer

Correct option: A.

$43$

Point of intersection is obtained by solving the equation of parabola $y^2 = x$ and equation of line $2y = x,$ we have
$y^2 = x$ and $2y = x$
$\Rightarrow y^2 = 2y$
$\Rightarrow y^2 - 2y = 0$
$\Rightarrow y = 0$ or $y = 2$
$\Rightarrow x = 0$ or $x = 4$
Thus $O(0, 0)$ and $A(4, 2)$ are the points of intersection of the curve and straight line. Area bound by then
$\text{A}=\int\limits_0^4(\text{y}_1-\text{y}_2)\text{ dx}$
$\Big[\text{Where, y}_1 =\sqrt{\text{x}}\text{ and y}_2=\frac{\text{x}}{2}\Big]$
$=\int\limits_0^4\Big(\sqrt{\text{x}}-\frac{\text{x}}{2}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}-\frac{1}{2}\times\frac{\text{x}^2}{2}\Bigg]_0^4$
$=\Big[\frac{2}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^2}{4}\Big]_0^4$
$=\frac{2}{3}4^\frac{3}{2}-\frac{1}{4}\times4^2-0$
$=\frac{2}{3}\times2^3-\frac{16}{4}$
$=\frac{16}{3}-4$
$=\frac{16-12}{3}$
$=\frac{4}{3}\text{ Sq units}$

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MCQ 211 Mark

The area bounded by the curve $y^2= 8x$ and $x^2 = 8y$ is:

✓
$\frac{16}{3}\text{ sq. units}$
B
$\frac{3}{16}\text{ sq. units}$
C
$\frac{14}{3}\text{ sq. units}$
D
$\frac{3}{14}\text{ sq. units}$

Answer

Correct option: A.

$\frac{16}{3}\text{ sq. units}$

Point of intersection of both the parabolas $y^2 = 8x$ and $x^2 = 8y$ is obtaining by solving the two equations,
$\text{y}^{2} = \text{8x}$ and $\text{x}^{2} = \text{8y}$
$\therefore \frac{\text{y}^{4}}{64} - \text{8y} = 0$
$\Rightarrow \text{y}(\text{y}^{3} - 8^{3}) - 0$
$\Rightarrow \text{y} = 0$ or $\text{y} = 8$
$\Rightarrow \text{x} = 0$ or $\text{x} = 8$
$\therefore O(0, 0)$ and $A(8, 8)$ are the points of intersection.
Area of the shaded region $= \int\limits^{8}_{0} |\text{y}_{2} - \text{y}_{1}| \text{dx}$
$=\int\limits^{8}_{0} (\text{y}_{2} - \text{y}_{1}) \text{dx}$
$=\int\limits^{8}_{0} \big(\sqrt{8\text{x}} - \frac{\text{x}^{2}}{8}\big) \text{dx}$
$= \bigg[\frac{\sqrt{8}}{\frac{3}{2}} \text{x}^{\frac{3}{2}} - \frac{1}{8} \times \frac{\text{x}^{2}}{3}\bigg]^{8}_{0}$
$ = \frac{2}{3} \times \sqrt{8} \times 8^{\frac{3}{2}} - \frac{1}{8} \times \frac{8^{3}}{3} - 0$
$= \frac{2}{3} \times \sqrt{8} \times 8 \sqrt{8} - \frac{8^{2}}{3}$
$= \frac{2}{3} \times 8^{2} - \frac{8^{2}}{3}$
$=\frac{8^{2}}{3} (2 - 1)$
$=\frac{64}{3} \text{sq units}$

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MCQ 221 Mark

The area bounded by the parabola $y^2 = 4ax$ and $x^2 = 4ay$ is:

A
$\frac{8\text{a}^3}{3}$
✓
$\frac{16\text{a}^2}{3}$
C
$\frac{32\text{a}^2}{3}$
D
$\frac{64\text{a}^2}{3}$

Answer

Correct option: B.

$\frac{16\text{a}^2}{3}$

To find the point of intersection of the parabola substitute $\text{y} = \frac{\text{x}^{2}}{4\text{a}}$ in $y^2= 4ax$
We get,
$\frac{\text{x}^4}{16\text{a}^{2}}=4\text{ax}$
$\Rightarrow x^4 - 64a^3 x = 0$
$\Rightarrow x(x^3- 64a^3) = 0$
$\Rightarrow x = 0$ or $x = 4a$
$\Rightarrow y = 0$ or $y = 4a$
Therefore, the required area $\text{ABCD},$
$\text{A} =\int\limits^\text{4a}_0(\text{y}_1-\text{y}_2)\text{dx}\Big($Where$, \text{ y}_1 = 2\sqrt{\text{ax}}$ and $\text{ y}_2=\frac{\text{x}^2}{\text{4a}}\Big)$
$= \int\limits^\text{4a}_0\Big(2\sqrt{\text{ax}}-\frac{\text{x}^2}{\text{4a}}\Big)\text{dx}$
$=\bigg[\frac{4\sqrt{\text{a}}}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^{3}}{12\text{a}}\bigg]^\text{4a}_0$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}(\text{4a}^\frac{3}{2})-\frac{(\text{4a})^3}{\text{12a}}\bigg]-\bigg[\frac{4\sqrt{\text{a}}}{3}(0)^\frac{3}{2}-\frac{(0)^3}{\text{12a}}\bigg]$ $$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}\text{8a}^\frac{3}{2}-\frac{64\text{a}^3}{12\text{a}}\bigg]-0$
$= \frac{32\text{a}^2}{3}-\frac{16\text{a}^3}{3}$
$= \frac{16\text{a}^2}{3}$ square units

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MCQ 231 Mark

The area bounded by the curve $x = 3y^2 – 9$ and the line $x = 0, y = 0$ and $y = 1$ is:

✓
$8\text{ sq.}\text{units}$
B
$\frac{8}{3}\text{ sq.}\text{units}$
C
$\frac{3}{8}\text{ sq.}\text{units}$
D
$3\text{ sq.}\text{units}$

Answer

Correct option: A.

$8\text{ sq.}\text{units}$

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MCQ 241 Mark

Area lying between the curves $y^2 = 4x$ and $y = 2x$ is:

A
$\frac{2}{3}$
✓
$\frac{1}{3}$
C
$\frac{1}{4}$
D
$\frac{3}{4}$

Answer

Correct option: B.

$\frac{1}{3}$

The points of intersection of the straight line and the parabola is obtained by solving the simultaneous equations,
$y^2 = 4x$ and $y = 2x$
$\Rightarrow (2x)^2 = 4x$
$\Rightarrow 4x^2 = 4x$
$\Rightarrow x(x - 1) = 0$
$\Rightarrow x = 0$ or $x = 1$
$\Rightarrow y = 0$ or $y = 2$
Thus, $O(0, 0)$ and $A(1, 2)$ are the points of intersection of the parabola and straight line shaded area is the required area.
Using the horizontal strip method, shaded area
$= \int\limits^2_0|\text{x}_2-\text{x}_1|\text{dy}$
$=\int\limits^2_0\Big[\Big(\frac{\text{y}}{2}\Big)-\Big(\frac{\text{y}^2}{4}\Big)\Big]\text{dy}$
$=\Big[\frac{1}{2}\Big(\frac{\text{y}^2}{2}\Big)-\frac{1}{4}\Big(\frac{\text{y}^3}{3}\Big)\Big]^2_0$
$=\frac{1}{4}(2)^2-\frac{1}{12}(2^3)-0$
$= 1 -\frac{8}{12}$
$= \frac{12-8}{12}$
$=\frac{1}{3}\text{ sq. units}$

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MCQ 251 Mark

The area bounded by the curve $\text{y}=\cos\text{x}$ in one are of the curve is where $=4\text{n}+1,\text{x}\in \text{integer:}$

A
$2\text{a}$
✓
$\frac{1}{\text{a}} $
C
$\frac{2}{\text{a}}$
D
$2{\text{a}^2}$

Answer

Correct option: B.

$\frac{1}{\text{a}} $

$=\text{Area} =\int\limits^\frac{\pi}{2}_0\cos\text{ax}\text{ dx}$
$=\Big[\frac{\sin\text{ax}}{\text{x}}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{0}-0=\frac{1}{\text{a}}$

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MCQ 261 Mark

The area bounded by the curve $y^2= 8x,$ the $x-$axis and the lastus rectum is:

✓
$\frac{16}{3}$
B
$\frac{23}{3}$
C
$\frac{32}{3}$
D
$\frac{16\sqrt{2}}{3}$

Answer

Correct option: A.

$\frac{16}{3}$

$y^2 = 8x$ represents a parabola opening side ways,
with vertex at $O(0, 0)$ and focus at $B(2, 0)$
Thus $AA\ '$ represents the latus rectum of the parabola.
The points of intersection of the parabola and latus rectum are $A(2, 4)$ and $A\ '(2, -4)$
Area bound by curve $, x-$axis and latus return is the area $\text{OABO},$
The approximating rectangle of with $= dx$ and length $= y$ has area $= y dx,$ and moves from $x = 0$ to
$\text{x} = 2$ area$\text{ OABO}= \int\limits^2_0|\text{y}|\text{dx}$
$= \int\limits^2_0\text{y}\text{ dx} $ $\{\text{y}>0, \Rightarrow|\text{y}|=\text{y}\}$
$= \int\limits^2_0\sqrt{8\text{x}}\text{ dx}$
$= 2\sqrt{2}\int\limits^2_0\sqrt{\text{xdx}}$
$= 2\sqrt{2}\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}\Bigg]^2_0$
$=2 \sqrt{2}\times \frac{2}{3}\Big(2^\frac{3}{2}-0\Big)$
$= 4 \frac{\sqrt{2}}{3}\times2\sqrt{2}$
$=\frac{16}{3}\text{ sq. units}$

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MCQ 271 Mark

Area of the region bounded by the curve $y^2 = 4x, y-$axis and the line $y = 3$, is:

A
$2$
✓
$\frac{9}{4}$
C
$\frac{9}{3}$
D
$\frac{9}{2}$

Answer

Correct option: B.

$\frac{9}{4}$

$y^2 = 4x$ represents a parabola with vertex at origin $O(0, 0)$ and symmetric about $+ve\ x-$axis
$y = 3$ is a straight line parallel to the $x-$axis
Point of intersection of the line and the parabola is given by
substituting $y = 3$ in the equation of the parabola
$y^2 = 4x$
$\Rightarrow 3^2 = 4x$
$\Rightarrow\text{x}=\frac{9}{4}$
Thus$, \text{A}=\Big(\frac{9}{4},3\Big)$ is the point of intersection of the parabola and straight line.
Required area is the shaded area $\text{OABO}$
Using the horizontal strip method,
Area $\text{(OABO)}= \int\limits^3_0|\text{x}|\text{ dy}$
$= \int\limits^3_0\frac{\text{y}^2}{4}\text{dy}$
$=\Big[\frac{1}{4}\Big(\frac{\text{y}^2}{3}\Big)\Big]^3_0$
$= \frac{3^3}{12}$
$=\frac{9}{4}\text{ sq. units}$

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MCQ 281 Mark

The area of the region bounded by $y = | x – 1 |$ and $y = 1$ is:

A
$2$
✓
$1$
C
$\frac{1}{2}$
D
$\frac{1}{4}$

Answer

Correct option: B.

$1$

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MCQ 291 Mark

The area of the region bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates $\text{x}=0,\text{x}=\frac{\pi}{2}$ and the $x-$axis is:

A
$\text{2 sq. units}$
B
$\text{4 sq. units}$
C
$\text{3 sq. units}$
✓
$\text{1 sq. unit}$

Answer

Correct option: D.

$\text{1 sq. unit}$

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MCQ 301 Mark

The area bounded by the curve $x^2 = 4y$ and straight line $x = 4y - 2$ is:

A
$\frac{3}{8}$
B
$\frac{5}{8}$
C
$\frac{7}{8}$
✓
$\frac{9}{8}$

Answer

Correct option: D.

$\frac{9}{8}$

The area bounded by the curve, $x^2 = 4y$, and line, $x = 4y - 2$, is represented by the shaded area $\text{OBAO}.$
Let $A$ and $B$ be the points of intersection of the line and parabola.
Coordinates of point $A$ are $\Big(-1,\frac{1}{4}\Big)$
Coordinates of point $B$ are $(2,1).$
We draw $AL$ and $BM$ perpendicular to $x-$axis.
It can be observed that, Area $\text{OBAO} =$ Area $\text{OBCO} +$ Area $\text{OACO} ...(1)$
Then, Area $\text{OBCO} =$ Area $\text{OMBC} -$ Area $\text{OMBO}$
$=\int\limits^2_0\frac{\text{x}+2}{4}\text{dx}-\int\limits^2_0\frac{\text{x}^2}{4}\text{dx}$
$=\frac{1}{4}\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^2_0-\frac{1}{4}\Big[\frac{\text{x}^3}{3}\Big]^2_0$
$=\frac{1}{4}[2+4]-\frac{1}{4}\Big[\frac{8}{3}\Big]$
$=\frac{3}{2}-\frac{2}{3}=\frac{5}{6}$
Similarly, Area $\text{OACO} =$ Area $\text{OLAC} -$ Area $\text{OLAO}$
$=\int\limits^0_{-1}\frac{\text{x}+2}{4}\text{dx}-\int\limits^0_{-1}\frac{\text{x}^2}{4}\text{dx}$
$=\frac{1}{4}\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^0_{-1}\frac{1}{4}\Big[\frac{\text{x}^3}{3}\Big]^0_{-1}$
$=-\frac{1}{4}\Big[\frac{(-1)^2}{2}+2(-1)\Big]-\Big[-\frac{1}{4}\Big(\frac{(-1)^3}{3}\Big)\Big]$
$=-\frac{1}{4}\Big[\frac{1}{2}-2\Big]-\frac{1}{12}$
$=\frac{1}{2}-\frac{1}{8}-\frac{1}{12}=\frac{7}{24}$
Therefore, required area, $=\Big(\frac{5}{6}+\frac{7}{24}\Big)=\frac{9}{8}\text{ sq}.\text{ units}$

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MCQ 311 Mark

The area bounded by the curve $y = x^4 - 2x^3 + x^2 + 3$ with $x-$axis and ordinates corresponding to the minima of $y$ is:

A
$1$
B
$\frac{91}{30}$
C
$\frac{30}{9}$
✓
$4$

Answer

Correct option: D.

$4$

Clearly, from the figure the minimum value of $y$ is $3$ when $x=0$ or $1.$
Therefore, the required area $\text{ABCD},$
$\text{A} = \int\limits^1_0\text{y}\text{ dx} ($Where, $y = x^4 - 2x^3 + x^2 + 3)$
$= \int\limits^1_0(\text{x}^4-2\text{x}^3+\text{x}^2+3)\text{dx}$
$=\bigg[\frac{\text{x}^5}{5}-\frac{2\text{(x})^4}{4}+\frac{\text{x}^3}{3}+3\text{x}\bigg]^1_0$
$=\bigg[\frac{(1)^5}{5}-\frac{2(1)^4}{4}+\frac{(1)^3}{3}+3(1)\bigg]-\bigg[\frac{(0)^5}{5}-\frac{2(0)^4}{4}+\frac{(0)^3}{3}+3(0)\bigg]$
$=\big[\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3\big]- 0$
$=\frac{6-15+10+90}{3}$
$=\frac{91}{30}\text{ square units}$

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MCQ 321 Mark

The area bounded by $\text{f(x)}=\text{x}^2,0\leq\text{x}\leq1,\text{g(x)}=\text{x}+2,1\leq\text{x}\leq2$ and $x –$ axis is:

A
$\frac{3}{2}$
B
$\frac{4}{3}$
C
$\frac{8}{3}$
✓
None of these

Answer

Correct option: D.

None of these

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MCQ 331 Mark

The ratio of the areas between the curves $\text{y}=\cos\text{x}$ and $\text{y}=\cos2\text{x}$ and x-axis from x = 0 to x = 0 to $\text{x}=\frac{\pi}{3}$

A
$1:2$
B
$2:1$
C
$\sqrt{3}:1$
✓
none of these

Answer

Correct option: D.

none of these

The line $\text{x} = \pi3$ meets the curve $\text{y} = \cos\text {x}\text{ at}\text { B}\pi3,12$

Area between the curve y = cos x and x - axis from x = 0 and $\text{x} = 3\pi$ is,

$=\Big[2-\frac{1}{2}-\frac{1}{3}\Big]-\Big[-4-2+\frac{8}{3}\Big]$

$= 2 -\frac{1}{2}-\frac{1}{3}+4+2-\frac{8}{3}$

$=8-\frac{1}{2}-\frac{9}{3}$

$=5-\frac{1}{2}$

$=\frac{9}{2}\text{ square units}$

The line $\text{x}=\frac{\pi}{3}$ meets the curve y = cos 2x at $\text{B}'\pi3, -12$ Area between the curve y = cos 2x and x -axis from x = 0 and $\text{x}=\frac{\pi}{3}$ is,

$= \text{A}_2 = \int\limits^\frac{\pi}{4}_0\text{y}_2\text{ dx}-\int\limits^\frac{\pi}{3}_\frac{\pi}{4}\text{y}_2\text{dx}$ $\big[\text{where}, \text{y}_2 = \cos(2\text{x})\big]$

$=\int\limits^\frac{\pi}{4}_0\cos(2\text{x})\text{dx}-\int\limits^\frac{\pi}{3}_\frac{\pi}{4}\cos(2\text{x})\text{ dx}$

$=\Big[\frac{1}{2}\sin(2\text{x})\Big]^\frac{\pi}{4}_0-\Big[\frac{1}{2}\sin(2\text{x})\Big]^\frac{\pi}{3}_\frac{\pi}{4}$

$=\frac{1}{2}\Big[\sin\Big(\frac{\pi}{2}\Big)-\sin(0)\Big]-\frac{1}{2}\Big[\sin\Big(\frac{2\pi}{3}\Big)-\sin\Big]$

$= \frac{1}{2}-\frac{1}{2}\Big[\frac{\sqrt{3}}{2}-1\Big]$

$= \frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{1}{2}$

$= 1-\frac{\sqrt{3}}{4}$

$=\frac{4-\sqrt{3}}{4}$

Therefore the retios will be

$\text{A}_1:\text{A}_2=\frac{\text{A}_1}{\text{A}_2}=\frac{\frac{\sqrt{3}}{2}}{\frac{4-\sqrt{3}}{4}}=\frac{2\sqrt{3}}{4-\sqrt{3}}$

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MCQ 341 Mark

The area enclosed by the curves $y^2 = x$ and $y = |x|$ is:

A
$\frac{2}{3}$
B
$1$
✓
$\frac{1}{6}$
D
$\frac{1}{3}$

Answer

Correct option: C.

$\frac{1}{6}$

Required area $=\text{A}=\int\limits^1_0\big(\sqrt{\text{x}}-\text{x}\big)\text{dx}$
$=\Big[\frac{2}{3}\text{x}^{\frac{3}{2}}-\frac{\text{x}^2}{2}\Big]^1_0$
$=\frac{2}{3}-\frac{1}{2}$
$=\frac{1}{6}$

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MCQ 351 Mark

The area bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates $\text{x}=0,\text{x}=\pi$ and the x-axis is:

✓
$2\text{ sq. units}$
B
$4\text{ sq. units}$
C
$3\text{ sq. units}$
D
$1\text{ sq. units}$

Answer

Correct option: A.

$2\text{ sq. units}$

$\text{A}=\int^\limits{\pi}_0\text{y}\text{ dx}$

$=\int^\limits{\pi}_0\sin(\text{x})\text{dx}$

$=\big[-\cos(\text{x})\big]^{\pi}_0$

$= -\cos(\pi)+\cos(0)$

$= 1 + 1$

$= 2 \text{ square units}$

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MCQ 361 Mark

Choose the correct answer from the given four options:
The area of the region bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates x = 0, $\text{x}=\frac{\pi}{2}$ and the x-axis is:

A
$2\text{ sq. units}$
B
$4\text{ sq. units}$
C
$3\text{ sq. units}$
✓
$1\text{ sq. units}$

Answer

Correct option: D.

$1\text{ sq. units}$

Area of the region bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates x = 0, $\text{x}=\frac{\pi}{2}$ and the X-axis is

$\text{A}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x dx}$

$=-\Big[\cos\text{x}\Big]^{\frac{\pi}{2}}_0=-\Big[\cos\frac{\pi}{2}-\cos0\Big]$

$=-[0-1]=1\text{ sq. units}$

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MCQ 371 Mark

For the area bounded by the curve $y = ax,$ the line $x = 2$ and $x -$ axis to be $2 \ sq.$ units, the value of a must be equal to:

A
$2$
✓
$4$
C
$6$
D
$8$

Answer

Correct option: B.

$4$

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MCQ 381 Mark

The area bounded by the curve y = f(x), x-axis, and the ordinates x = 1 and $(\text{b}-1)\sin(3\text{b}+4)$ Then, f(x) is:

A
$(\text{x}-1)\cos(3\text{x}+4)$
B
$\sin(3\text{x}+4)$
✓
$\sin(3\text{x}+4)+3(\text{x}-1)\cos(3\text{x}+4)$
D
none of these

Answer

Correct option: C.

$\sin(3\text{x}+4)+3(\text{x}-1)\cos(3\text{x}+4)$

sin (3x + 4) + 3 (x - 1) cos (3x + 4)
y = fx

If A is the area bound by the curve, x-axis, x = 1 and x = b

$\Rightarrow \int\limits^\text{b}_1\text{f}(\text{x})\text{dx}=\big[\text{A}\big]^\text{b}_1 = (\text{b - 1})\sin(3\text{b} + 4)$ {given}

$\Rightarrow \text{f}(\text{x})=\frac{\text{d}}{\text{dx}}((\text{x} - 1)\sin(3\text{x}+ 4))$

$= \sin (3\text{x} + 4)\frac{\text{d}}{\text{dx}}(\text{x} - 1) + (\text{x} - 1)\frac{\text{d}}{\text{dx}}\sin(3\text{x} + 4)$

$= \sin (3\text{x} + 4) + 3(\text{x} - 1)\cos(3\text{x} + 4)$

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MCQ 391 Mark

Area bounded by the curve $\text{y}=\sin\text{x}$ and the x-axis between $\text{x}=0$ and $\text{x}=2\pi$ is:

A
2 sq units
B
0 sq units
C
3 sq units
✓
4 sq units

Answer

Correct option: D.

4 sq units

(d), as $\text{x}=\sin$ is positive in 1st and 2nd quadrant and negative is 3rd and 4th quadrant.

$=\text{Area}=\int\limits^{2\pi}_0\sin\text{x}\text{ dx}$

$=\int\limits^\pi_0\sin\text{x}+\int\limits^{2\pi}_\pi(-\sin\text{x})\text{dx}$

$=4\text{sq}\text{ units}$

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MCQ 401 Mark

The area of the region formed by $\text{x}^2+\text{y}^2-6\text{x}-4\text{y}+12\leq0,\text{ y}\leq\text{x}$ and $\text{x}\leq\frac{5}{2}$

A
$\frac{\pi}{6}-\frac{\sqrt{3}+1}{8}$
B
$\frac{\pi}{6}+\frac{\sqrt{3}+1}{8}$
✓
$\frac{\pi}{6}-\frac{\sqrt{3}-1}{8}$
D
none of these

Answer

Correct option: C.

$\frac{\pi}{6}-\frac{\sqrt{3}-1}{8}$

We have,
$\text{x}^{2} + \text{y}^{2}-6\text{x}-4\text{y}+\leq0$
$\text{y}\leq\text{x}$
$\text{x}\leq\frac{5}{2}$
Following are the corresponding equations of the given inequation.
$x^2 + y^2 -6x -4y +12 = 0 ...(i)$
$y = x ...(ii)$
$\text{x} = \frac{5}{2}\ ...(\text{iii})$
Here, $\text{ABC}$ is our required region in which point $A$ is intersection of $(i)$ and $(iii),$ point $B$ is intersection of $(i)$ and $(ii)$ and point $C$ is intersection of $(ii)$ and $(iii)$ By solving $ (i), (ii)$ and $(iii)$ we get the coordinates of $B$ and $C$ as $B = (2, 2) \text{C} = \Big(\frac{5}{2}, \frac{5}{2}\Big)$ Now, the equation of the circle is,
$x^2 + y^2 - 6x - 4y +12 = 0$
$\Rightarrow (x-3)^2 + (y-2)^2 = 1$
$\Rightarrow (y - 2)^2 = 1 - (x - 3)^2$
$\Rightarrow\text{y}-2 = \pm\sqrt{1-(\text{x}-3)^{2}}$
$\Rightarrow\text{y} = \pm\sqrt{1-(\text{x}-3})^{2}+2$
$\Rightarrow\text{y} = \sqrt{1-(\text{x}-3)^{2}}+2 \text{ or }-\sqrt{1-(\text{x}{-3})^{2}}+2$
$\text{y}= \sqrt{1-(\text{x}-3)^{2}}+2$ is not possible,
Therefore $\text {y}= -\sqrt{1-(\text{x}-3)^{2}}+2$
The area of the required region $\text{ABC}$,
$\text{A}= \int\limits^\frac{5}{2}_{2}(\text{y}_2-\text{y}_1)\text{ dx} \big(\text{where},\text{ y}_1 = -\sqrt{1-(\text{x}-3)^2} +2\text{ and}\text{ y}^2 = \text{x}\big)$
$= \int\limits^\frac{5}{2}_{2}\big[\text {x}-\big(-\sqrt{1(\text{x}-3)^{2}}+2\big)\big]\text{dx}$
$= \int^\limits\frac{5}{2}_2\big[\text{x}+\sqrt{1-(\text{x}-3)^{2}}\big]\text{dx}$
$= \bigg[\frac{\text{x}^{2}}{2}+\frac{(\text{x}-3)}{2}\sqrt{1-(\text{x}-3)^{2}}+\frac{1}{2}\sin^{-1}(\text{x}-3)-2\text{x}\bigg]^{\frac{5}{2}}_{2}$
$=\Bigg[\frac{\big(\frac{5}{2}\big)^{2}}{2}+\frac{\frac{5}{2}-3}{2}\sqrt{1-\Big\{\Big(\frac{5}{2}\Big)-3\Big\}^2}+\frac{1}{2}\sin^{-1}=\Big(\frac{5}{2}-3\Big)-2\Big(\frac{5}{2}\Big)\Bigg]\\-\bigg[\frac{2^{2}}{2}+\frac{2-3}{2}\sqrt{1-(2-3)^2}+\frac{1}{2}\sin^{-1}(2-3)-2(2)\bigg]$
$= \bigg[\frac{25}{8}-\frac{1}{4}\sqrt{1-\frac{1}{4}}+\frac{1}{2}\sin^{-1}\Big(-\frac{1}{2}\Big)-5\bigg]\\- \bigg[2-\frac{1}{2}\times0+\frac{1}{2}\sin^{-1}(-1)-4\bigg]$
$= \Big[-\frac{15}{8}-\frac{\sqrt{3}}{8}+\frac{1}{2}\times\Big(-\frac{\pi}{6}\Big)\Big]-\Big[+\frac{1}{2}\times\Big(-\frac{\pi}{2}\Big)-2\Big]$
$= -\frac{15}{8}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}+2$
$=\frac{\pi}{6}-\frac{\sqrt{3}-1}{8}$

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MCQ 411 Mark

Choose the correct answer from the given four options:The area of the region bounded by the curve $x^2 = 4y$ and the straight line $x = 4y - 2$ is:

A
$\frac{3}{8}\text{ sq. units}$
B
$\frac{5}{8}\text{ sq. units}$
C
$\frac{7}{8}\text{ sq. units}$
✓
$\frac{9}{8}\text{ sq. units}$

Answer

Correct option: D.

$\frac{9}{8}\text{ sq. units}$

We have parabola $x^2 - 4y$ and the straight line $x = 4y - 2$
Solving we get
$x^2 = x + 2$
$\Rightarrow x^2 - x - 2 = 0$
$\Rightarrow (x - 2)(x + 1) = 0$
$\Rightarrow x = -1, 2$
For $x = -1, \text{y}=\frac{1}{4}$
and for $x = 2, y = 1$
Thus point of intersection are $\Big(-1,\frac{1}{4}\Big)$ and $(2,1)$
Grapha of parabola $x^2 = 4y$ and $x = 4y - 2$ are as show in the following figure.

$\therefore$ From the figure, area of shaded region
$\text{A}=\int\limits^2_{-1}\Big(\frac{\text{x}+2}{4}-\frac{\text{x}^2}{4}\Big)\text{dx}$
$=\frac{1}{4}\Big[\frac{\text{x}^2}{2}+2\text{x}-\frac{\text{x}^3}{3}\Big]^2_{-1}$
$=\frac{1}{4}\bigg[\Big(\frac{4}{2}+4-\frac{8}{3}\Big)-\Big(\frac{1}{2}-2+\frac{1}{3}\Big)\bigg]$ $=\frac{1}{4}\bigg[8-\frac{1}{2}-3\bigg]=\frac{9}{8}\text{ sq. units}$

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MCQ 421 Mark

The area bounded by the lines $y = |x – 2|, x = 1, x = 3$ and the $x-$ axis is:

✓
$1 \ sq.$ unit
B
$2 \ sq.$ units
C
$3 \ sq.$ units
D
$4 \ sq.$ units

Answer

Correct option: A.

$1 \ sq.$ unit

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MCQ 431 Mark

Choose the correct answer from the given four options:The area of the region bounded by parabola $y^2 = x$ and the straight line $2y = x$ is:

✓
$\frac{4}{3}\text{ sq. units}$
B
$1\text{ sq. units}$
C
$\frac{2}{3}\text{ sq. units}$
D
$\frac{1}{3}\text{ sq. units}$

Answer

Correct option: A.

$\frac{4}{3}\text{ sq. units}$

Solving $y^2 = x$ and $2y = x$, we get
$\Big(\frac{\text{x}}{2}\Big)^2=\text{x}$
$\Rightarrow\ \text{x}^2=4^{\frac{3}{2}}$
$\Rightarrow\ \text{x(x}-4)=0$
$\Rightarrow\ \text{x}=4,0$
When $x = 0, y = 0$ and when $x = 4, y = 2$
So, the intersection points are $(0, 0)$ and $(4, 2).$

Thus required area of shaded region,
$\text{A}=\int\limits^4_0\Big[\sqrt{\text{x}}-\frac{\text{x}}{2}\Big]\text{dx}$
$=\Bigg[\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2}\cdot\frac{\text{x}^2}{2}\Bigg]^4_0=\bigg[2\cdot\frac{\text{x}^{\frac{3}{2}}}{3}-\frac{\text{x}^2}{4}\bigg]^4 _0$
$=\frac{2}{3}4^{\frac{3}{2}}-\frac{16}{4}\frac{2}{3}\cdot+\frac{1}{4}\cdot0$
$=\frac{16}{3}-\frac{32}{12}=\frac{48-32}{12}$
$=\frac{16}{12}=\frac{4}{3}\text{ sq. units}$

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MCQ 441 Mark

The area bounded by the lines $y = |x| - 1$ and $y = -|x| + 1$ is:

A
$\text{1 sq. unit}$
✓
$\text{2 sq. unit}$
C
$2\sqrt{2}\text{ sq}.\text{units}$
D
$\text{4 sq. units}$

Answer

Correct option: B.

$\text{2 sq. unit}$

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MCQ 451 Mark

Area of the region bounded by the curve $\text{y}=\cos\text{x}$ between $x = 0$ and $\text{x}=\pi$ is:

✓
$2 \ sq$. units
B
$4 \ sq.$ units
C
$3 \ sq.$ units
D
$1 \ sq.$ units

Answer

Correct option: A.

$2 \ sq$. units

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MCQ 461 Mark

The area bounded by the curvey $=\sqrt{\text{x}}$ the line 2y + 3 = x and the x - axis in the first quadrant is:

✓
$9$
B
$\frac{27}{4}$
C
$36$
D
$18$

Answer

Correct option: A.

$9$

Given curves are $\text{y}=\sqrt{\text{x}}$ ...(1)and 2y - x + 3 = 0 ...(2)

Solving (1) and (2), we get

$=\sqrt{2}-(\sqrt{\text{x}})^2+3=0$

$\Rightarrow(\sqrt{\text{x}})^2-2\sqrt{\text{x}}-3=0$

$\Rightarrow(\sqrt{\text{x}}-3)(\sqrt{\text{x}}-3=0$

$\Rightarrow\sqrt{\text{x}}-3$

$\because\sqrt{\text{x}}=-1 \text{ is}\text{ not}\text{ possible}$

$\therefore\text{y}=3$

Hence required area

$=\int\limits^3_0(\text{x}_2-\text{x}_1\text{dy}$

$=\int\limits^3_0((2\text{y}+3)-\text{y}^2)\text{dy}$

$=\Big[\text{y}^2+3\text{y}-\frac{\text{y}^3}{3}\Big]^3_0$

$=9+9-9=9$

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MCQ 471 Mark

The area bounded by the line $y = 2x – 2, y = – x$ and $x-$axis is given by:

A
$\frac{9}{2}\text{ sq.}\text{ units}$
B
$\frac{43}{6}\text{ sq.}\text{ units}$
C
$\frac{35}{6}\text{ sq.}\text{ units}$
✓
None of these

Answer

Correct option: D.

None of these

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MCQ 481 Mark

The area bounded by $y - 1 = |x|, y = 0$ and $|x| =\frac{1}{2}$ will be:

A
$\frac{3}{4}$
B
$\frac{3}{2}$
✓
$\frac{5}{4}$
D
None of these

Answer

Correct option: C.

$\frac{5}{4}$

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MCQ 491 Mark

Area lying first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the line $x = 0$ and $x = 2,$ is:

✓
$\pi$
B
$\frac{\pi}{2}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{4}$

Answer

Correct option: A.

$\pi$

$x^2 + y^2 = 4$ represents a circle with centre at origion $O(0, 0)$ and radius $2$ units,
cutting the coordinate axis at $A, A\ ', B$ and $B\ ', x =2$
represents a straight line parallel to the $y-$axis,
intersecting the circle at $A(2, 0)x = 0$ respresents the $y-$axis
Area bounded by the circle and the two given lines in the first quadrant is the shaded area $\text{OBCAO}$
Area$\text{(OBCAO)}=\int\limits^2_0|\text{y}|\text{dx}$
$=\int\limits^2_0\sqrt{4-\text{x}^2}\text{dx}$
$=\bigg[\frac{1}{2}\text{x}\sqrt{4-\text{x}^2}+\frac{1}{2}\times4\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\bigg]^2_0$
$= \frac{1}{2}\times2\sqrt{4-2^2}+\frac{1}{2}\times4\sin^{-1}\Big(\frac{2}{2}\Big)-0$
$= 0 + 2 \sin^{-1}(1)$
$= 2\times\frac{\pi}{2}$
$= \pi\text{ sq. units}$

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MCQ 501 Mark

Choose the correct answer from the given four options:
The area of the region bounded by the y-axis, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x},0\leq\text{x}\leq\frac{\pi}{2}$ is:

A
$\sqrt{2}\text{ sq. units}$
B
$\big(\sqrt{2}+1)\text{ sq. units}$
✓
$\big(\sqrt{2}-1)\text{ sq. units}$
D
$\big(2\sqrt{2}-1)\text{ sq. units}$

Answer

Correct option: C.

$\big(\sqrt{2}-1)\text{ sq. units}$

We have, Y-axis i.e., x = 0, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x},$ where $0\leq\text{x}\leq\frac{\pi}{2}$

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MCQ 511 Mark

The area enclosed between the curve $\text{y}=\log_{\text{e}}(\text{x}+\text{e}),\text{x}=\log_\text{e}\Big(\frac{1}{\text{y}}\Big)$ and the $x-$ axis is:

✓
$2$
B
$1$
C
$4$
D
none of these

Answer

Correct option: A.

$2$

The point of intersection of the curves $\text{y}=\log_{\text{e}}(\text{x}+\text{e})$ and $\text{x}=\log_\text{e}\Big(\frac{1}{\text{y}}\Big) $
$\text{y}=\log_{\text{e}}(\text{x}+\text{e})$
$\Rightarrow\text{x}+\text{e}=\text{e}^{\text{y}}$
$\Rightarrow\text{x}=\text{e}^{\text{y}}-\text{e}$
and $\text{x}_{2} = \log_\text{e}\Big(\frac{1}{\text{y}}\Big)$
Therefore, area of the required region,
$\text{A} = \int\limits^1_0(\text{x}_2-\text{x}_1)\text{ dy} \Big[\text{where}, = \text{x}_1 = \text{e}^{\text{y}}-\text{e}\text{ and }\text{x}_2= \log_\text{e}\Big(\frac{1}{\text{y}}\Big)\Big]$
$\text{A} = \int\limits^1_0\log_\text{e}\Big(\frac{1}{\text{y}}\Big)\text{ dy}-\int\limits^1_0(\text{e}^\text{y}-\text{e})\text{ dy}$
$\text{A}=\int\limits^1_0\log_\text{e}\Big(\frac{1}{\text{y}}\Big)\text{ dy}-\big[\text{e}^{\text{y}}-\text{ey}\big]\ ...(\text{i})$
Let $\text{I} = \int\log_\text{e}\Big(\frac{1}{\text{y}}\Big)\text{dy}$
Putting $\frac{1}{\text{y}}=\text{t}$
Therefore, integral becomes
$\text{I} = \int-\frac{1}{\text{t}^{2}}\log_\text{e}\text{t}\text{ dt} $
$= -\log_\text{e}\text{t}\int\frac{1}{\text{t}^{2}}\text{ dt}-\int\frac{1}{\text{t}}\times\frac{1}{\text{t}}\text{ dt}$
$= \frac{1}{\text{t}}\log_\text{e}\text{t}+\frac{1}{\text{t}}$
$= \text{y}\log_\text{e}\frac{1}{\text{y}}+\text{y}$
Now, $(i)$ becomes
$\text{A} = \Big[\text{y}\log_\text{e}\frac{1}{\text{y}}+\text{y}\Big]^1_0-\big[\text{e}^\text{y}-\text{ey}\big]^1_0$
$=\Big[\text{y}\log_\text{e}\Big(\frac{1}{\text{y}}\Big)+\text{y}-\text{e}^{\text{y}}+\text{ey}\Big]^1_0$
$=\big[\log_\text{e}(1)+1-\text{e}^1+\text{e}(1)\big]-\big[0+0-\text{e}^{(0)}+\text{e}(0)\big]$
$= 2$

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MCQ 521 Mark

Compute the area of the figure bounded by straight lines $x = 0, x = 2$ and the curves $y = 2^x$ and $y = 2x - x^2$:

✓
$\frac{3}{\log 2}-\frac{4}{3}$
B
$\frac{3}{\log 2}+\frac{4}{3}$
C
$\frac{4}{\log 3}-\frac{4}{3}$
D
$\frac{4}{\log 2}+\frac{1}{3}$

Answer

Correct option: A.

$\frac{3}{\log 2}-\frac{4}{3}$

$=\int\limits^2_02^\text{x}-2\text{x}+\text{x}^2\text{ dx}$
$=\Big[\frac{2^\text{x}}{\text{In(2)}}+\frac{(\text{x}-3)\text{x}^2}{3}\Big]^2_0$
$=\Big[\frac{4}{\text{In}2}-\frac{4}{3}\Big]-\Big[\frac{1}{\text{In}2}\Big]$
$=\frac{3}{\text{log}\ 2}-\frac{4}{3}$

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MCQ 531 Mark

The area bounded by the curve $y = f(x),$ the $x-$axis and $x = 1$ and $x = b$ is $(b – 1)$ $\sin(3b + 4).$ Then$, f(x)$ is:

A
$(\text{x}-1)\cos(3\text{x}+4)$
B
$\cos(3\text{x}+4)$
✓
$\sin(\text{3x}-4)+3(\text{x}-1).\cos(3\text{x}+4)$
D
None of these

Answer

Correct option: C.

$\sin(\text{3x}-4)+3(\text{x}-1).\cos(3\text{x}+4)$

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MCQ 541 Mark

Choose the correct answer:Area of the region bounded by the curve $y^2 = 4x, y-$axis and the line $y = 3$ is:

A
$2$
✓
$\frac94$
C
$\frac93$
D
$\frac92.$

Answer

Correct option: B.

$\frac94$

The equation of curve is $y^2 = 4x$
We are to find the area bounded by the curve $y^2 = 4x, y-$axis and the line $y = 3.$
Required area $=\int\limits^3_0\text{x dy}=\int\limits^3_0\frac{\text{y}^2}{4}\text{dy}$
$=\frac14\int\limits^3_0\text{y}^2\text{dy}=\frac14\Big[\frac{\text{y}^3}{3}\Big]^3_0$
$=\frac{1}{12}\Big[\text{y}^3\Big]^3_0=\frac{1}{12}(27-0)$
$=\frac{27}{12}=\frac94$

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MCQ 551 Mark

The area of the region bounded by the curve $=2x - x^2$ and the line $y = x$ is $........$ square units:

✓
$\frac{1}{6}$
B
$\frac{1}{2}$
C
$\frac{1}{3}$
D
$\frac{7}{6}$

Answer

Correct option: A.

$\frac{1}{6}$

We note that the region bounded by these curves is in the region $x \in [0, 1].$
In this region, the curve $y = x$ lies below the curve $y = 2x - x^2$ So, to
calculate the area of said region, we evaluatie the following integral:
$=\int\limits^1_02\text{x}-\text{x}^2-\text{x}\text{dx}$
$=\int\limits^1_0\text{x}-\text{x}^2\text{dx}$
$=\Big[\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}\Big]^\text{x=1}_{\text{x=0}}$
$=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.$

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MCQ 561 Mark

Area of the region bounded by the curve $y = |x + 1| + 1, x = –3, x = 3$ and $y = 0$ is:

A
$8 \ sq$ units
✓
$16 \ sq$ units
C
$32 \ sq$ units
D
None of these

Answer

Correct option: B.

$16 \ sq$ units

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MCQ 571 Mark

The area $($in $sq.$ units$)$ bounded by the curves $\text{y}=\sqrt{\text{x}},2\text{y}-\text{x}+3=0$ and $x-$axis lying in the first quadrant is:

✓
$9$
B
$36$
C
$18$
D
$\frac{27}{4}$

Answer

Correct option: A.

$9$

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MCQ 581 Mark

The area common to the parabola $y = 2x^2$ and $y = x^2+ 4$ is:

A
$\frac{2}{3}\text{ sq. units}$
B
$\frac{3}{2}\text{ sq. units}$
✓
$\frac{32}{3}\text{ sq. units}$
D
$\frac{3}{32}\text{ sq. units}$

Answer

Correct option: C.

$\frac{32}{3}\text{ sq. units}$

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MCQ 591 Mark

The area of the region bounded by the curve $x^2 = 4y$ and the straight line $x = 4y - 2$ is:

A
$\frac{3}{8}\ \text{sq}.\text{units}$
B
$\frac{5}{8}\ \text{sq}.\text{units}$
C
$\frac{7}{8}\ \text{sq}.\text{units}$
✓
$\frac{9}{8}\ \text{sq}.\text{units}$

Answer

Correct option: D.

$\frac{9}{8}\ \text{sq}.\text{units}$

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MCQ 601 Mark

The value of $aa$ for which the area between the curves $y^2 = 4ax$ and $x^2 = 4ay$ is $\text{1 sq.unit,}$ is:

A
$\sqrt{3}$
B
$4$
C
$4\sqrt{3}$
✓
$\frac{\sqrt{3}}{4}$

Answer

Correct option: D.

$\frac{\sqrt{3}}{4}$

$=\text{y}^2=4\text{ ax}$
$=\text{y}=\sqrt{4\text{ ax}}$
$=\text{x}^2=4\text{ ax}$
$=\text{y}=\frac{\text{x}^2}{4\text{a}}$
$=$area$=\int\limits^{4\text{a}}_0\sqrt{4\text{a}\text{x}}\text {d}\text{x}-\int\limits^{4\text{a}}_0\frac{\text{x}^2}{4\text{a}}\text{dx}$
$=2\sqrt{\text{a}}\times\frac{2}{3}\text{(x})^\frac{3}{2}\Big]^{4\text{a}}_0-\frac{\text{x}^3}{3(4\text{a})}\Big]^{4\text{a}}_0$
$=\frac{32\text{a}^2}{3}-\frac{16\text{a}^2}{3}$
$=\frac{16\text{a}^2}{3}=1$
$=\text{a}=\frac{\sqrt{3}}{4}$

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MCQ 611 Mark

The area of the region bounded by the ellipse $\frac{\text{x}\ ^2}{25}+\frac{\text{y}^2}{16}=1$ is.

✓
$25\pi\text{ sq.}\text{ units}$
B
$20\pi^2\text{ sq.}\text{ units}$
C
$16\pi^2\text{ sq.}\text{ units}$
D
$25\pi\text{ sq.}\text{ units}$

Answer

Correct option: A.

$25\pi\text{ sq.}\text{ units}$

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MCQ 621 Mark

The area under the curve $y = x^4$ and the lines $x = 1, x = 5$ and $x-$axis is:

A
$\frac{3124}{3}\text{ sq.}\text{ units}$
B
$\frac{3124}{7}\text{ sq.}\text{ units}$
✓
$\frac{3124}{5}\text{ sq.}\text{ units}$
D
$\frac{3124}{9}\text{ sq.}\text{ units}$

Answer

Correct option: C.

$\frac{3124}{5}\text{ sq.}\text{ units}$

Concept:
The area under the function $y = f(x)$ from $x = a$ to $x = b$ and the $x-$axis is given by the definite integral
$\int\limits^\text{b}_\text{a}\text{f(x)}\text{dx}$
This is for curves that are entirely on the same side of the $x-$axis in the given range.
If the curves are on both sides of the $x-$axis, then we calculate the areas of both sides separately and add them.
Definite integral:
If$\int\text{f(x)}\text{dx}=\text{g(x)}+\text{c},$then
$\int\limits^\text{b}_\text{a}\text{f(x)}\text{dx}=[\text{g(x)}]^\text{b}_\text{a}=\text{g(b)}-\text{g(a)}$
$\int\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n+1}}{\text{n+1}}+\text{c} $
Calculation:
$\int\text{x}^4\text{dx}=\frac{\text{x}^5}{5}+\text{c.}$
Using the above concept for area of a curve, we can say that the required area is:
$\text{I}=\int\limits^5_1\text{x}^4\text{dx}$
$=\Big[\frac{\text{x}^5}{5}\Big]^5_1$
$=\frac{5^5}{5}-\frac{1^5}{5}$
$=\frac{3125-1}{5}$
$=\frac{3124}{5}$

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MCQ 631 Mark

The area bounded by the curve $y^2= x$, line $y = 4$ and $y-$axis is:

A
$\frac{16}{3}\ \text{sq.}\text{units}$
✓
$\frac{64}{3}\ \text{sq.}\text{units}$
C
$7\sqrt{2}\ \text{sq.}\text{units}$
D
None of these

Answer

Correct option: B.

$\frac{64}{3}\ \text{sq.}\text{units}$

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MCQ 641 Mark

The area common to the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ and $\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{a}^2}=1,0<\text{b}<\text{a}$ is:

A
$(\text{a}+\text{b})^2\tan^{-1}\frac{\text{b}}{\text{a}}$
B
$(\text{a}+\text{b})^2\tan^{-1}\frac{\text{a}}{\text{b}}$
✓
$4\text{a}+\text{b}\tan^{-1}\frac{\text{b}}{\text{a}}$
D
$4\text{a}+\text{b}\tan^{-1}\frac{\text{a}}{\text{b}}$

Answer

Correct option: C.

$4\text{a}+\text{b}\tan^{-1}\frac{\text{b}}{\text{a}}$

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MCQ 651 Mark

Area bounded by the curvey $\text{y}=\text{x}+\sin\text{x}$ and its inverse function between the ordinates $\text{x}=0$ and $\text{x}=2\pi$ is:

A
$8\pi\text{ sqp}.\text{units}$
B
$4\pi\text{ sq}.\text{units}$
✓
$8\pi\text{ sq}.\text{units}$
D
$3\pi\text{ sq}.\text{units}$

Answer

Correct option: C.

$8\pi\text{ sq}.\text{units}$

Inverse function is the mirror image with respect to y = x
Then area bounded by $\text{x}+\sin\text{x}$ and its inverse function is
$=4\int\limits^\pi_0(\text{x}+\sin\text{x}-\text{x})\text{ dx}=8$

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MCQ 661 Mark

The area enclosed between the graph of $y = x^3$ and the lines $x = 0, y = 1, y = 8$ is:

✓
$\frac{45}{4}$
B
$14$
C
$7$
D
None of these

Answer

Correct option: A.

$\frac{45}{4}$

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MCQ 671 Mark

Area bounded between the curve $x^2 = y$ and the line $y = 4x$ is:

✓
$\frac{32}{3}\text{sq}\text{ unit}$
B
$\frac{1}{3}\text{sq}\text{ unit}$
C
$\frac{8}{3}\text{sq}\text{ unit}$
D
$\frac{16}{3}\text{sq}\text{ unit}$

Answer

Correct option: A.

$\frac{32}{3}\text{sq}\text{ unit}$

Given curves are $x^2 = y$ and $y = 4x$
Intersection points are $(0, 0)$ and $(4, 16)$
$\therefore$ Required area
$=\int\limits^4_0(4\text{x}-\text{x}^2)\text{dx}$
$=\Big[\frac{4\text{x}^2}{2}-\frac{\text{x}^3}{3}\Big]^4_0$
$=\Big[32-\frac{64}{3}\Big]$
$=\frac{32}{3}\text{sq}\text{ unit}$

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MCQ 681 Mark

Area enclosed by the circle $x^2 + y^2 = a^2$ is equal to:

A
$2\pi\text{a}^2\text{sq.}\text{ units}$
✓
$\pi\text{a}^2\text{sq.}\text{ units}$
C
$2\pi\text{a}\text{ sq.}\text{ units}$
D
$\pi\text{a}\text{ sq.}\text{ units}$

Answer

Correct option: B.

$\pi\text{a}^2\text{sq.}\text{ units}$

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MCQ 691 Mark

The area of the region bounded by the curves $= xe^x, y = xe^{−x}$ and the line $x=1$ is:

A
$\frac{4}{\text{e}}$
B
$\frac{3}{\text{e}}$
✓
$\frac{2}{\text{e}}$
D
$\frac{1}{\text{e}}$

Answer

Correct option: C.

$\frac{2}{\text{e}}$

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MCQ 701 Mark

The area of the region bounded by the parabola $y = x^2 + 1$ and the staight line $x + y = 3$ is given by:

A
$\frac{45}{7}$
B
$\frac{25}{4}$
C
$\frac{\pi}{18}$
✓
$\frac{9}{2}$

Answer

Correct option: D.

$\frac{9}{2}$

To find the point of intersection of the parabola
$y = x^2 + 1$ and the line $x + y = 3$
substitute $y = 3 - x$ in $y = x^2 + 13 - x = x^2 + 1$
$\Rightarrow x^2 + x - 2 = 0$
$\Rightarrow (x - 1)(x + 2) = 0$
$\Rightarrow x = 1$ or $x = -2$
$\therefore y = 2$ or $y = 5$
So, we get the points of intersection $A (-2, 5)$ and $C (1, 2).$
Therefore, the required area $\text{ABC},$
$\text{A} = \int\limits^1_{-2}(\text{y}_1-\text{y}_2)\text{dx}$ $\big(\text{Where}, \text{y}_1 = 3-\text{x }\text{and}\text{ y}_2 = \text{x}^2+1\big)$
$=\int\limits^1_{-2}\big[(3-\text{x})-(\text{x}^2+1)\big]\text{dx}$
$=\int\limits^1_{-2}(3-\text{x}-\text{x}^2-1)\text{dx}$
$=\int\limits^1_{-2}\big(2-\text{x}-\text{x}^2\big)\text{dx}$
$= \Big[2\text{x}-\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}\Big]^1_{-2}$
$=\bigg[2(1)-\frac{(1)^2}{2}-\frac{(1)^3}{3}\bigg]-\bigg[2(-2)-\frac{(-2)^2}{2}-\frac{(-2)^3}{3}\bigg]$
$=\Big[2-\frac{1}{2}-\frac{1}{3}\Big]-\Big[-4-2+\frac{8}{3}\Big]$
$=2-\frac{1}{2}-\frac{1}{3}+4+2-\frac{8}{3}$
$=8-\frac{1}{2}-\frac{9}{3}$
$=5-\frac{1}{2}$
$=\frac{9}{2}\text{ square units}$

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MCQ 711 Mark

The area of the region $\{(\text{x},\text{y}):\text{y}^2\leq4\text{x},4\text{x}^2+4\text{y}^2\leq9\}$ is:

✓
$\frac{\sqrt2}{6}+\frac{9\pi}{8}-\frac{9}{4}\sin-1(\frac{1}{5})$
B
$\frac{\sqrt2}{6}-\frac{9\pi}{8}$
C
$\frac{9\pi}{8}-\frac{9}{4}\sin-1(\frac{1}{3})$
D
None of these

Answer

Correct option: A.

$\frac{\sqrt2}{6}+\frac{9\pi}{8}-\frac{9}{4}\sin-1(\frac{1}{5})$

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MCQ 721 Mark

The area bounded by $y = x^2$ and $y = 1 - x^2$ is:

✓
$\frac{\sqrt{8}}{3}$
B
$\frac{16}{3}$
C
$\frac{32}{3}$
D
$\frac{17}{3}$

Answer

Correct option: A.

$\frac{\sqrt{8}}{3}$

Required are
$=2\Bigg[\int\limits^\frac{1}{\sqrt{2}}_0(1+\text{x}^2)\text{dx}-\int\limits^\frac{1}{\sqrt{2}}_0\text{x}^2\text{dx}\Bigg]$
$=2\Bigg[\text{x}+\frac{\text{x}^3}{3}\Bigg]^\frac{1}{\sqrt{2}}_0-2\Bigg[\frac{\text{x}^3}{3}\Bigg]^\frac{1}{\sqrt{2}}_0$
$=\frac{\sqrt{8}}{3}$

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MCQ 731 Mark

The area bounded by $y = x^2, y = [x+1], \text{x}\leq1$ and the $y -$ axis is:

A
$\frac{1}{3}$
✓
$\frac{2}{3}$
C
$1$
D
$\frac{7}{3}$

Answer

Correct option: B.

$\frac{2}{3}$

Required area
$=\int\limits^1_0\sqrt{\text{y}}\text{dx}$
$=\frac{2}{3}$

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MCQ 741 Mark

Area between the parabola $x^2 = 4y$ and line $x = 4y –2$ is:

A
$\frac{8}{9}$
B
$\frac{9}{7}$
C
$\frac{7}{9}$
✓
$\frac{9}{8}$

Answer

Correct option: D.

$\frac{9}{8}$

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MCQ 751 Mark

If the curves $y = x^3 + ax$ and $y = bx^2 + c$ pass through the point $(-1, 0)$ and have common tangent line at this point, then the value of $a+b$ is?

A
$0$
✓
$-2$
C
$-3$
D
$-1$

Answer

Correct option: B.

$-2$

As the curve pass through the point $P(-1, 0) 0 = a = -1$
$\Rightarrow a = -10 = b +$ Common tangent at this point
$=\frac{\text{dx}}{\text{dx}}=2\text{ bx} $ and $\frac{\text{dy}}{\text{dx}}=3\text{x}^2+\text{a}$
$= 3 - 1 = 2$
$2bx = 2 - 2b =2b $
$= -1a + b = -2$

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MCQ 761 Mark

The area bounded by the curves $\text{y}=\sin\text{x},\text{y}=\cos\text{x}$ y − axes in first quadrant is:

✓
$\sqrt{2}-1$
B
$\sqrt{2}$
C
$\sqrt{2}+1$
D
$\text{none}\text{ of}\text{ the}\text{ above}$

Answer

Correct option: A.

$\sqrt{2}-1$

The area bounded by the curves $\text{y}=\sin\text{x},\text{y}=\cos\text{x}$ y − axes in first quadrant is,

$\text{A}=\int\limits^\frac{\pi}{4}_0(\cos\text{x}-\sin\text{x})\text{dx}$

$=[\sin\text{x}+\cos\text{x}]^\frac{\pi}{4}_0$

$=\Big(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\Big)-(\sin0+\cos0)$

$=\sqrt{2}-1$

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MCQ 771 Mark

The area bounded by $y –1 = |x|, y = 0$ and $|x| \frac{1}{2}$ will be:

A
$\frac{3}{4}$
B
$\frac{3}{2}$
✓
$\frac{5}{4}$
D
None of these

Answer

Correct option: C.

$\frac{5}{4}$

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MCQ 781 Mark

Choose the correct answer from the given four options:
The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = -1 is:

A
$4\text{ sq. units}$
B
$\frac{3}{2}\text{ sq. units}$
✓
$6\text{ sq. units}$
D
$8\text{ sq. units}$

Answer

Correct option: C.

$6\text{ sq. units}$

Required area, $\text{A}=\int\limits^1_{-1}(2\text{y}+3)\text{dy}$

$=\Big[\frac{2\text{y}^2}{2}+3\text{y}\Big]^1_{-1}$

$\Big[\text{y}^2+3\text{y}\Big]^1_{-1}$

$=\big[1+3-1+3\big]$

$=6\text{ sq. units}$

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MCQ 791 Mark

The area bounded by the curve $x^2+ y^2= 1$ in first quadrant is:

✓
$\frac{\pi}{4}\ \text{sq.}\text{units}$
B
$\frac{\pi}{2}\ \text{sq.}\text{units}$
C
$\frac{\pi}{3}\ \text{sq.}\text{units}$
D
$\frac{\pi}{6}\ \text{sq.}\text{units}$

Answer

Correct option: A.

$\frac{\pi}{4}\ \text{sq.}\text{units}$

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MCQ 801 Mark

The area bounded by the curve $y = (x + 1)^2, y = (x - 1)^2$ and the line $y = 0$ is:

A
$\frac{1}{6}$
✓
$\frac{2}{3}$
C
$\frac{1}{4}$
D
$\frac{1}{3}$

Answer

Correct option: B.

$\frac{2}{3}$

$\text{R.E.F}$ image $\rightarrow y = (x + 1)^2$ is obtained by shifting origin to $(-1, 0)$ in $x^2= y$,
for $y = (x - 1)^2$ Similarly $(1, 0)$ As graph is symmetric about $y -$ axis, area $A$ would be,
$=\text{A}=2\int\limits^1_0(\text{x}-1)^2\text{dx}$
$=2\int\limits^1_0\text{x}^2-2\text{x}+1\text{dx}=2$
$\Big[\text{x}\frac{3}{3}-\text{x}^2+\text{x}\Big]^1_0$
$=2\Big(\frac{1}{3}-1+1\Big)=\frac{2}{3}$

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MCQ 811 Mark

The area bounded by the parabola $y^2= 4ax$, latus rectum and $x-$axis is:

A
$0$
✓
$\frac{4}{3}\text{a}^2$
C
$\frac{2}{3}\text{a}^2$
D
$\frac{\text{a}^2}{3}$

Answer

Correct option: B.

$\frac{4}{3}\text{a}^2$

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MCQ 821 Mark

Points of inflexion of the curve $y = x^4 - 6x^3+ 12x^2 + 5x + 7$ are

A
$(1, 19); (1, 12)$
✓
$(1, 19); (2, 33)$
C
$(1, 2); (2, 1)$
D
$(1, 7); (2, 6)$

Answer

Correct option: B.

$(1, 19); (2, 33)$

$y = x^4 - 6x^3 + 12x^2 + 5x + 7y(x)$
$= 4x^3 - 18x^2 + 24x + 5y(x)$
$= 12x^2- 36x + 24y(x)$
$= 012x^2 - 36x + 24 = 0x^2 - 3x + 2$
$= 0x^2 - 2x - x + 2$
$= 0x(x - 2) -1(x - 2) = 0(x - 1) -1(x - 2)$
$= 0(x - 1)(x - 2) = 0(x - 1)(x - 2) = 0x = 1, 2$
Inflection point of a function is where the function changes from concave up to concave down or vice $-$ versa $x < 1, f(x) > 01 < x < 2, f(x) < 0x > 2, f(x) > 0$
$\because f(x)$ changes sign
$\therefore$ At $x = 1, 2y = f(x)$ has inflection point At $x = 1, y = f(1) = 19$ At $x = 2, x = f(2) = 33x = 2, y = f(2) = 33$
Point of inflection $(1,19); (2,33)$

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MCQ 831 Mark

Area between the parabolas $y^2 = 4ax$ and $x^2 = 4ay$ is:

A
$\frac{2}{3}\text{a}^2-5$
B
$\frac{15}{4}\text{a}^2+5$
C
$\frac{16}{3}\text{a}^2+2$
✓
$\frac{16}{3}\text{a}^2$

Answer

Correct option: D.

$\frac{16}{3}\text{a}^2$

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MCQ 841 Mark

The area bounded by \displaystyle $y = xe^{∣X∣}$ and $|x| = 1$ is:

A
$4$
B
$6$
C
$1$
✓
$2$

Answer

Correct option: D.

$2$

$=\text{I}=\int\limits^1_{-1}\text{y}\text{dx}$
$=\int\limits^1_{-1}\text{dx}^\text{x}\text{dx}$
$=\int\limits^1_{-1}\text{dx}^\text{-x}\text{dx}+\int\limits^1_0\text{xe}^\text{x}\text{dx}$
$=[-\text{xe}^\text{-x}+\int\text{e}^\text{-x}\text{ dx}]^0_{-1}+[\text{xe}^\text{x}-\int\text{e}^\text{x}\text{dx}]^1_0$
$=[-\text{xe}^\text{-x}-\text{e}^\text{-x}]^0_{-1}+[\text{xe}^\text{x}-\text{e}^\text{x}]^1_0$
$=-1-\text{(e}-\text{e})]+[\text{e}-\text{e}(-1)]=|-1|+|1|$
we take modulus because area can not be negative and this function is symmetry about $y$ axis.
we have to put modulus otherwise area will be zeroso,ans is $2$

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MCQ 851 Mark

Choose the correct answer in the following.
The area bounded by the curve y = x|x|, x-axis and the ordinates x = -1 and x = 1 is given by,

A
0
B
$\frac13$
✓
$\frac23$
D
$\frac43.$

Answer

Correct option: C.

$\frac23$

$\text{Required area}=\int\limits^1_0\text{y dx}$
$=\int\limits^1_{-1}\text{x}|\text{x}|\text{dx}$
$=\int\limits^{0}_{-1}\text{x}^2\text{dx}+\int\limits^1_0\text{x}^2\text{dx}$
$=\Big[\frac{\text{x}^3}{3}\Big]^0_{-1}+\Big[\frac{\text{x}^3}{3}\Big]^1_0$
$=-\Big(-\frac13\Big)+\frac13$
$=\frac23\text{ units}$
Thus, the correct answer is C.

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MCQ 861 Mark

The area under the curve $y= 2x^3+ 4x^2$ between $x = 2, x = 4$ is:

✓
$192.6$
B
$198.6$
C
$88.3$
D
$172.3$

Answer

Correct option: A.

$192.6$

The area under the curve is given as
$=\int\limits^4_2\ 2\text{x}^3+4\text{x}^2\text{ dx}$
$=\int\limits^4_2\ 2\text{x}^3\text{ dx}+4\text{x}^2\text{ dx}$
$=2\frac{\text{x}^4}{4}\Big|^4_2+4\frac{\text{x}^3}{3}\Big|^\frac{2}{4}_2$
$=64\times2-8+\frac{4^4}{3}-\frac{32}{3}$
$=120-\frac{224}{3}=192.6$

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MCQ 871 Mark

The area of the plane region bounded by the curves $x + 2y^2 = 0$ and $x + 3y^2 = 1$ is equal to:

A
$\frac{5}{3}$
B
$\frac{1}{3}$
C
$\frac{2}{3}$
✓
$\frac{4}{3}$

Answer

Correct option: D.

$\frac{4}{3}$

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MCQ 881 Mark

The area bounded by curve $y = x^2- 1$ and tangents to it at $(2, 3)$ and $y -$ axis is:

A
$\frac{8}{3}$
✓
$\frac{2}{3}$
C
$\frac{4}{3}$
D
$\frac{1}{3}$

Answer

Correct option: B.

$\frac{2}{3}$

$=\text{x}-$axis$:(-1,0)$
$=$Area$=\int\limits^0_{-1}(\text{x}^2-1)\text{dx}$
$=\Big[\frac{\text{x}^3}{3}-\text{x}\Big]^0_{-1}$
$=\Big[\frac{-1}{3}-(-1)\Big]-[0]$
$=-\frac{1}{3}+1$
$=\frac{2}{3}\text{sq}.\text{ units}$

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MCQ 891 Mark

The area of the region enclosed by the lines $y = x, x = e$ and curve $\text{y}=\frac{1}{\text{x}}$ and the positive $x -$ axis is:

A
$1\text{ sq.}\text{ units}$
✓
$\frac{3}{2}\text{ sq.}\text{ units}$
C
$\frac{5}{2}\text{ sq.}\text{ units}$
D
$\frac{1}{2}\text{ sq.}\text{ units}$

Answer

Correct option: B.

$\frac{3}{2}\text{ sq.}\text{ units}$

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MCQ 901 Mark

A rea bounded by the circle $x^2 + y^2 = 1$ and the curve $| x | + | y | = 1$ is:

A
$2\pi$
✓
$\pi-2$
C
$\pi$
D
$\pi+3$

Answer

Correct option: B.

$\pi-2$

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MCQ 911 Mark

Choose the correct answer from the given four options : The area of the region bounded by the circle $x^2 + y^2 = 1$ is :

A
$2\pi\text{ sq. units}$
✓
$\pi\text{ sq. units}$
C
$3\pi\text{ sq. units}$
D
$4\pi\text{ sq. units}$

Answer

Correct option: B.

$\pi\text{ sq. units}$

Here, $x^2 + y^2 = 1^2$ is a circle with centre at $(0, 0)$
$\Rightarrow\ \text{y}^2=1-\text{x}^2$
$\Rightarrow\ \text{y}=\sqrt{1=\text{x}^2}$
Graph for the circle $x^2 + y^2 = 1^2$ is shown below:

$\therefore$ Area enclosed by circle $=2\int\limits^{1}_{-1}\sqrt{1^2-\text{x}^2}\text{dx}$
$=2.2\int\limits^{1}_{0}\sqrt{1^2-\text{x}^2}\text{dx}$
$=2\cdot2\bigg[\frac{\text{x}}{2}\sqrt{1^2-\text{x}^2}+\frac{1^2}{2}\sin^{-1}\frac{\text{x}}{1}\bigg]^{1}_{0} $
$=4\bigg[\frac{1}{2}\cdot0+\frac{1}{2}\cdot\frac{\pi}{2}-0-\frac{1}{2}\cdot0\bigg] $
$=4\cdot\frac{\pi}{4}=\pi\text{ sq. units} $

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MCQ 921 Mark

Find area bounded by curves $\{(\text{x},\text{y}):\text{y}\geq\text{x}^2$ and ${y}=\text{x}\}$ :

A
$\frac{5}{3}$
B
$\frac{1}{2}$
✓
$\frac{1}{3}$
D
$\frac{1}{9}$

Answer

Correct option: C.

$\frac{1}{3}$

$=\text{y}=\text{x}=\{\text{x};\text{x}\geq0-\text{x};\text{x}<0 <0\}\text{p}$ and $Q$ are $x^2 = x$
$= x^2 - x = 0\ x(x - 1) = 0\ x = 0,1Q =1$ similarlyp
$=-\text{A}=\int\limits^1_0\text{x}-\text{x}^2\text{ dx}$
$=\text{A}=\Big[\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}\Big]^1_0$
$\text{A}=\frac{1}{2}-\frac{1}{3}$
$=\text{A}=\frac{1}{3}$

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MCQ 931 Mark

The area of the region enclosed by the parabola $x^2 = y,$ the line $y = x + 2$ and the $x -$ axis, is:

A
$\frac{2}{9}$
✓
$\frac{9}{2}$
C
$9$
D
$2$

Answer

Correct option: B.

$\frac{9}{2}$

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MCQ 941 Mark

Choose the correct answer : Area lying between the curves $y^2 = 4x$ and $y = 2x$ is:

A
$\frac23$
✓
$\frac13$
C
$\frac14$
D
$\frac34.$

Answer

Correct option: B.

$\frac13$

Equation of curve $($parabola$)$ is $y^2 = 4x ...(i)$

$\Rightarrow\text{y}=2\sqrt{\text{x}}=2\text{x}^{\frac12}...(\text{ii})$
Equation of another curve $($line$)$ is $y = 2x ...(iii)$ Solving eq. $(i)$ and $(iii),$
we get $x = 0$ or $x = 1$ and $y = 0$ or $y = 2$
Therefore, Points of intersections of circle $(i)$ and line $(ii)$ are $O(0, 0)$ and $A(1, 2).$
Now Area $\text{OBAM}=$ Area bounded by parabola $(i)$ and $x-$ axis
$=\Bigg|\int\limits^1_0\text{y dx}\Bigg|=\Bigg|\int\limits^1_02\text{x}^{\frac12}\text{dx}\Bigg|=$ $2\frac{\Big(\text{x}^{\frac32}\Big)^1_0}{\frac32}$
$=\frac43(1-0)=\frac43\dots(\text{iv})$
Also, Area $\Delta\text{ OAM}=$ Area bounded by parabola $(iii) $and $x-$ axis
$=\Bigg|\int\limits^1_0\text{y dx}\Bigg|=\Bigg|\int\limits^1_02\text{x dx}\Bigg|=2\Big(\frac{\text{x}^2}{2}\Big)^1_0 = (1 - 0) = 1 ...(v)$
Now Required shaded area $\text{OBA} =$ Area $\text{OBAM} -$ Area of $\Delta\text{ OAM}$ $=\frac43-1=\frac{4-3}{3}=\frac13\text{ sq. units}$
Therefore, option $(B)$ is correct.

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MCQ 951 Mark

Consider the following statements:Statement $I:$ The area bounded by the curve$, \text{y}=\sin\text{x}$ between $\text{x}=0$ and $x = 2p$ is $\text{2 sq. units}.$Statement $II:$ The area bounded by the curve, $\text{y}=2\cos\text{x}$ and the $x-$axis from $\text{x}=0$ to $x = 2p$ is $\text{8 sq. units}.$

A
Statement $I$ is true
✓
Statement $II$ is true
C
Both statements are true
D
Both statements are false

Answer

Correct option: B.

Statement $II$ is true

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MCQ 961 Mark

The area of the region bounded by the and the lines $x = 2$ and $x = 3.$

✓
$\frac{7}{2}\text{ sq}.\text{units}$
B
$\frac{9}{2}\text{ sq}.\text{units}$
C
$\frac{11}{2}\text{ sq}.\text{units}$
D
$\frac{13}{2}\text{ sq}.\text{units}$

Answer

Correct option: A.

$\frac{7}{2}\text{ sq}.\text{units}$

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MCQ 971 Mark

Choose the correct answer in the following : The area of the circle $x^2 + y^2 = 16$ exterior to the parabola $y^2 = 6x$ is:

A
$\frac43(4\pi-\sqrt3)$
B
$\frac43(4\pi+\sqrt3)$
✓
$\frac43(8\pi-\sqrt3)$
D
$\frac43(8\pi+\sqrt3).$

Answer

Correct option: C.

$\frac43(8\pi-\sqrt3)$

The given equations are
$x^2 + y^2 = 16 ...(1)$
$y^2 = 6x ...(2)$

Area bounded by the circle and parabola
$= 2[$Area $\text{(OADO)} \ + $ Area $(\text{ADBA})]$
$=2\bigg[\int\limits^2_0\sqrt{6\text{x}}\text{ dx}+\int\limits^4_2\sqrt{16-\text{x}^2}\text{dx}\bigg]$
$=2\Bigg[\sqrt6\left\{\frac{\text{x}^{\frac32}}{\frac32}\right\}^2_0\Bigg]$ $+2\Big[\frac{\text{x}}{2}\sqrt{16-\text{x}^2}+\frac{16}{2}\sin^{-1}\frac{\text{x}}{4}\Big]^4_2$
$=2\sqrt6\times\frac23\Big[\text{x}^{\frac32}\Big]^2_0 +2\Big[8.\frac{\pi}{2}-\sqrt{16-4}-8\sin^{-1}\Big(\frac12\Big)\Big]$
$=\frac{4\sqrt6}{3}(2\sqrt2)+2\Big[4\pi-\sqrt{12}-8\frac{\pi}{6}\Big]$
$=\frac{16\sqrt3}{3}+8\pi-4\sqrt3-\frac83\pi$
$=\frac43\Big[4\sqrt3+6\pi-3\sqrt3-2\pi\Big]$
$=\frac43\Big[\sqrt3+4\pi\Big]$
$=\frac43\Big[4\pi+\sqrt3\Big]\text{ units}$
Area of circle $= n(r)^2$
$= n(4)^2$
$= 16n$ units
$\therefore$ Required area $=16\pi-\frac43\Big[4\pi+\sqrt3\Big]$
$=\frac43\Big[4\times3\pi-4\pi-\sqrt3\Big]$
$=\frac43\big(8\pi-\sqrt3\big)\text{ units}$
Thus, the correct answer is $C.$

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MCQ 981 Mark

If the area bounded by the x - axis, curve y = f (x) and the lines x = 1, x = b is equal to $\sqrt{\text{b}^2+1}-\sqrt{2}$ for all b > 1, then f(x) is:

A
$\sqrt{\text{x}-1}$
B
$\sqrt{\text{x}+1}$
C
$\sqrt{\text{x}^2+1}$
✓
$\frac{\text{x}}{\sqrt{1+\text{x}^2}}$

Answer

Correct option: D.

$\frac{\text{x}}{\sqrt{1+\text{x}^2}}$

$\text{Atp},\int\limits^\text{b}_1\text{f(x)}\text{ dx}=\sqrt{\text{b}^2+1}-\sqrt{2}\int\limits^\text{b}_1$
$\text{f(x)}\text{ dx}=\Big[\sqrt{\text{x}^2+1}\Big]^\text{b}_1$
$\text{f(x)}=\text{d}(\sqrt{\text{x}^2}+1)$
$\text{f(x)}=\frac{\text{x}}{\sqrt{1+\text{x}^2}}$

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MCQ 991 Mark

The area bounded by the parabola $x = 4 - y^2$ and $y-$ axis, in square units, is:

A
$\frac{3}{32}$
✓
$\frac{32}{3}$
C
$\frac{33}{2}$
D
$\frac{16}{3}$

Answer

Correct option: B.

$\frac{32}{3}$

The points of intersection of the parabola and the $y-$ axis are $A(0, 2)$ and $C(0, -2)$.
Therefore, the area of the required region $\text{ABCO},$
$\text{A} = \int\limits^2_{-2}\text{x }\text{dy}$
$= \int\limits^2_{-2}(4-\text{y}^{2})\text{dy}$
$= \Big[4\text{y}-\frac{\text{y}^{3}}{3}\Big]^2_{-2}$
$= \Big[4(2)-\frac{(2)^3}{3}\Big]-\Big[4(2)-\frac{(-2)^3}{3}\Big]$
$= \Big(8-\frac{8}{3}\Big)-\Big(-8+\frac{8}{3}\Big)$
$=8-\frac{8}{3}+8-\frac{8}{3}$
$= 16 -\frac{16}{3}$
$= \frac{32}{3}\text{ square units}$

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MCQ 1001 Mark

Choose the correct answer: Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is:

✓
$\pi$
B
$\frac{\pi}{2}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{4}.$

Answer

Correct option: A.

$\pi$

The equation of circle is $x^2 + y^2 = 4$
we are to find the area of the circle lying between the circle lying between the lines $x = 0$ and $x = 2$ in the first quadrant.
Required area $=\int\limits^2_0\text{y dx}$
$=\int\limits^2_0\sqrt{4-\text{x}^2}\text{dx}\ \ [\therefore\text{of}(1)]$

$=\int\limits^2_0\sqrt{(2)^2-\text{x}^2}\text{ dx} =\Big[\frac{\text{x}}{2}\sqrt{(2)^2-\text{x}^2}+\frac{(2)^2}{2}\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\Big]^2_0$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}+2\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\Big]^2_0$
$=\Big[\frac{\text{2}}{2}\sqrt{4-\text{4}}+2\sin^{-1}\Big(\frac{\text{2}}{2}\Big)\Big]-[0+2\sin^{-1}0]$
$=\Big[0+2\sin^{-1}(1)-[0+2\times0]\Big]$
$=2\sin^{-1}1=2\times\frac{\pi}{2}=\pi$

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MCQ 1011 Mark

Choose the correct answer from the given four options:
Area of the region bounded by the curve $\text{y}\cos\text{x}$ between x = 0 and $\text{x}=\pi$ is:

✓
$2\text{ sq. units}$
B
$4\text{ sq. units}$
C
$3\text{ sq. units}$
D
$1\text{ sq. units}$

Answer

Correct option: A.

$2\text{ sq. units}$

Required area enclosed by the curve $\text{y}\cos\text{x},$ and x = 0 and $\text{x}=\pi$

$\text{A}=\int\limits^{\frac{\pi}{2}}_0\cos\text{x dx}\Bigg|\int\limits^\pi_{\frac{\pi}{2}}\cos\text{x dx}\Bigg|$

$=\Big[\sin\frac{\pi}{2}-\sin0\Big]+\Big|\sin\frac{\pi}{2}-\sin\pi\Big|$

$=1+1=2\text{ sq. units}$

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MCQ 1021 Mark

If the area above the $x-$ axis, bounded by the curve $y = 2kx$ and $x = 0,$ and $x = 2$ is $\frac{3}{\log_{\text{e}}2},$ then the value of $k$ is :

A
$\frac{1}{2}$
✓
$1$
C
$-1$
D
$2$

Answer

Correct option: B.

$1$

The area bounded by the curves $y = 2^{kx}, x = 0,$ and $x = 2$ is given by $\int\limits^2_02^\text{kx}\text{dx}.$
It is given that $\int\limits^2_02^{\text{kx}}\text{dx} = \frac{3}{\log_{\text{e}}(2)} $
$\Rightarrow\frac{1}{\text{k}}\bigg[\frac{2^\text{kx}}{\log_e(2)}\bigg]^2_0=\frac{3}{\log_{e}(2)}$
$\Rightarrow\frac{1}{\text{k}}\bigg[\frac{2^\text{k(2)}}{\log_e(2)}-\frac{2^{\text{k(0)}}}{\log_{e}(2)}\bigg]= \frac{3}{\log_{e}(2)}$
$\Rightarrow\frac{1}{\text{k}}\Big(\frac{2^{\text{2k}}}{\log_e(2)}-\frac{1}{\log_e(2)}\Big)= \frac{3}{\log_e(2)}$
$\Rightarrow\frac{1}{\text{k}}(2^{\text{2k}}-1)=3$
$\Rightarrow(2^{\text{2k}}-1)=3\text{k}$
$\Rightarrow2^{\text{2k}}-3\text{k}-1=0$
$\Rightarrow \text{k}= 1$
Clearly, $K = 1$ satisfies the equation.
Hence, $K = 1$

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MCQ 1031 Mark

The area bounded by the curve $x^2 = 4y + 4$ and line $3x + 4y = 0$ is:

A
$\frac{25}{4}\text{sq}.\text{units}$
B
$\frac{125}{8}\text{sq}.\text{units} $
C
$\frac{125}{16}\text{sq}.\text{units}$
✓
$\frac{124}{4}\text{sq}.\text{units}$

Answer

Correct option: D.

$\frac{124}{4}\text{sq}.\text{units}$

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MCQ 1041 Mark

Area of the region bounded by the curve $\text{y}=\sqrt{49-\text{x}^2}$ and the x - axis is:

✓
$\frac{49}{2}\pi\text{ sq}\text{ units}$
B
$98\pi\text{ sq}\text{ units}$
C
$49\pi\text{ sq}\text{ units}$
D
$240\pi\text{ sq}\text{ units}$

Answer

Correct option: A.

$\frac{49}{2}\pi\text{ sq}\text{ units}$

$=\text{as}\text{ area}\text{ is}\text{ above}\text{ the}\text{ the}\text{x}-\text{axis}$
$\therefore\text{area}=2\int\limits^7_0\sqrt{49-\text{x}^2\text{dx}}$
$=2\Big[\frac{\text{x}}{2}\sqrt{49-\text{x}^2}+\frac{49}{2}\sin^{-1}\frac{\text{x}}{7}\Big]^7_0$
$=2\Big[\Big(\frac{7}{2}\times+\frac{49}{2}\sin^{-1}1\Big)-(0)\Big]$
$=\frac{49}{2}\pi\text{ sq}\text{ units}$

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MCQ 1051 Mark

The line $y = mx$ bisects the area enclosed by lines $\text{x}=0,\text{y}=0$ and $\text{x}=\frac{3}{2}$ and the curve $\text{y}=1+4\text{x}-\text{x}^2.$ Then the value of $m$ is:

✓
$\frac{13}{6}$
B
$\frac{13}{2}$
C
$\frac{13}{5}$
D
$\frac{13}{7}$

Answer

Correct option: A.

$\frac{13}{6}$

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MCQ 1061 Mark

The area of the region bounded by the parabola $y = x^2$ and $y = |x|$ is:

A
$3$
B
$\frac{1}{2}$
✓
$\frac{1}{3}$
D
$2$

Answer

Correct option: C.

$\frac{1}{3}$

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MCQ 1071 Mark

He area bounded by $y = x^2, x = y^2$ is:

✓
$1$
B
$\frac{1}{6}$
C
$\frac{3}{4}$
D
$\text{None}\text{ of}\text{ these}$

Answer

Correct option: A.

$1$

$=\text{y}=\text{x}^2,\text{y}^2=\text{x}$
$\Rightarrow\text{y}=\sqrt{\text{x}}$
The curves intersect at $(0, 0)$ and $(1,1)$ Area between the curves is given by
$=\int\limits^1_0\sqrt{\text{x}}-\text{x}^2\text{dx}$
$=\frac{1}{2}\text{x}^\frac{3}{2}+\frac{\text{x}^3}{3}\Big|^1_0$
$=\frac{2}{3}+\frac{1}{3}=1$

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MCQ 1081 Mark

The area bounded by $y = 2 - x^2$ and $x + y = 0$ is :

A
$\frac{7}{2}\text{ sq. units}$
✓
$\frac{9}{2}\text{ sq. units}$
C
$9\text{ sq. units}$
D
$\text{none of these}$

Answer

Correct option: B.

$\frac{9}{2}\text{ sq. units}$

$\text{(IMAGE)}$
To find the points of intersection of $x + y = 0$ and $y = 2 - x^2.$ We put $x = -y$ in $y = 2 - x^2,$
We get $y = 2 - y^2$
$\Rightarrow y^2 + y - 2 = 0$
$\Rightarrow y - 1, y + 2 = 0$
$\Rightarrow y = 1, -2$
$\Rightarrow x = -1, 2$
Therefore, the points of intersection are $A(-1, 1)$ and $C(2, -2).$ The area of the required region $\text{ABCD},$
$\text{A} = \int\limits^2_{-1}(\text{y}_1-\text{y}_{2})\text{dx} \ ($Where, $y_1 = 2 - x^2$ and $y_2 = -x)$
$=\int\limits^2_{-1}(2-\text{x}^{2}+\text{x})\text{dx}$
$ = \Big[2\text{x}-\frac{\text{x}^{3}}{3}+\frac{\text{x}^{2}}{2}\Big]^2_{-1}$
$= \bigg\{2(2)-\frac{(2)^{3}}{3}+\frac{(2)^{2}}{2}\bigg\}-\bigg\{2(-1)-\frac{(-1)^{3}}{3}+\frac{(-1)^{2}}{2}\bigg\}$
$= \Big(4-\frac{8}{3}+2\Big)-\Big(-2+\frac{1}{3}+\frac{1}{2}\Big)$
$=6-\frac{8}{3}+2-\frac{1}{3}-\frac{1}{2}$
$= 8 - \frac{9}{3}-\frac{1}{2}$
$= 5 -\frac{1}{2}$
$\frac{9}{2}\text{ sq. units}$

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MCQ 1091 Mark

The area between x-axis and curve $\text{y}=\cos\text{x}$ when $0\leq\text{x}\leq2\pi$ is:

A
0
B
2
C
3
✓
4

Answer

Correct option: D.

Required shaded area,

$\text{A} = \int\limits^\frac{\pi}{2}_0\cos\text{x}\text{ dx} + \int\limits^\frac{3\pi}{2}_\frac{\pi}{2}(-\cos\text{x})\text{dx} + \int\limits^{2\pi}_\frac{3\pi}{2}\cos\text{x}\text{ dx}$

$= \int\limits^\frac{\pi}{2}_0\cos\text{x}\text{ dx}-\int\limits^\frac{3\pi}{2}_\frac{\pi}{2}\cos\text{x}\text{ dx} + \int\limits^{2\pi}_\frac{3\pi}{2}\cos\text{x}\text{ dx}$

$=\Big[\sin\text{x}\Big]^{\frac{\pi}{2}}_0-\Big[\sin\text{x}\Big]^{\frac{3\pi}{2}}_\frac{\pi}{2}+\Big[\sin\text{x}\Big]^{2\pi}_\frac{3\pi}{2}$

$= \Big[\sin\text{x}\Big]^\frac{\pi}{2}_0-\Big[\sin\text{x}\Big]^\frac{3\pi}{2}_\frac{\pi}{2}+\Big[\sin\text{x}\Big]^{2\pi}_\frac{3\pi}{2}$

$=(1-0)-(-1-1)+\big[0-(-1)\big]$

$=1+2+1$

$=4\text{ sq. units}$

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MCQ 1101 Mark

Find the area above $x-$axis, bounded by the curves $y = 2^{kx}, x = 0$ and $x = 2$:

✓
$\frac{4^\text{k}-1}{\text{k}\text{ log}_\text{e}2}$
B
$\frac{2^\text{k}-1}{2\text{ log}_\text{e}2}$
C
$\frac{3-\text{k}}{\text{k}\text{ log}_\text{e}2}$
D
$\frac{-1+3^\text{k}}{2\text{ log}_\text{e}2}$

Answer

Correct option: A.

$\frac{4^\text{k}-1}{\text{k}\text{ log}_\text{e}2}$

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MCQ 1111 Mark

The area of the region bounded by the parabola $(y - 2)^2 = x - 1,$ the tangent to it at the point with the ordinate $3$ and the $x-$ axis is:

A
$3$
B
$6$
C
$7$
✓
none of these

Answer

Correct option: D.

none of these

The tangent passes through the point with ordinate $3,$ so substituting $y = 3$ in equation of parabola $(y - 2)^2 = x - 1,$ we get $x = 2$
Therefore, the line touches the parabola at $(2, 3)$
We have,
$(\text{y}-2)^{2} = \text{x}-1$
$\Rightarrow \text{y}-2 = \sqrt{\text{x}-1}$
$\Rightarrow \text{y} = \sqrt{\text{x}-1}+2$
Slope of tangent of parabola at $x = 2$
$\Big[\frac{\text{dy}}{\text{dt}}\Big]_{\text{x}=2}=\Big[\frac{1}{2\sqrt{\text{x}-1}}\Big]_{\text{x}=2}=\frac{1}{2}$
Therefore the equation of the tangent is given as:
$\text{y}-\text{y}_0 = \text{m}(\text{x}-\text{x}_0)$
$\Rightarrow\text{y}-3=\frac{1}{2}(\text{x}-2)$
$\Rightarrow\text{y}=\frac{1}{2}\text{x}+2$
Therefore, area of the required region ABC,
$\text{A} = \int\limits^3_0(\text{x}_1-\text{x}_2)\text{dy}$ $\big[\text{Where}, \text{x}_1=(\text{y}-2)^{2}+1\text{ and}\text{ x}_2=2(\text{y}-2)\big]$
$= \int\limits^3_0(\text{x}_1-\text{x}_2)\text{dy}$
$=\int\limits^3_0(\text{y}-2)^2+1-2(\text{y}-2)\text{dy}$
$= \int\limits^3_0\big[(\text{y}-2)-1\big]^2\text{dy}$
$= \int\limits^3_0\big[\text{y}-3\big]^2\text{dy} $
$= \bigg[\frac{(\text{y}-3)^3}{3}\bigg]^3_0$
$=\bigg[\frac{(3-3)^{3}}{3}\bigg]-\bigg[\frac{(0-3)^3}{3}\bigg]$
$=9$

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MCQ 1121 Mark

The area bounded by the curve $\text{y}^2=16\text{x}$ and line $\text{y}=\text{ mx} $ is $\frac{2}{3},$ then $m$ is equal to:

A
$3$
✓
$4$
C
$1$
D
$2$

Answer

Correct option: B.

$4$

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MCQ 1131 Mark

The area bounded by the lines $|x| + |y| = 1$ is:

A
$\text{1 sq. unit}$
✓
$\text{2 sq. units}$
C
$2\sqrt{2}\text{ sq}.\text{units}$
D
$\text{4 sq. units}$

Answer

Correct option: B.

$\text{2 sq. units}$

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MCQ 1141 Mark

Find the area enclosed by the parabola $y^2 = x$ and the line $y + x = 2$ and the $x-$axis:

A
$\frac{5}{6}\text{ sq.}\text{units}$
✓
$\frac{7}{6}\text{ sq.}\text{units}$
C
$\frac{6}{7}\text{ sq.}\text{units}$
D
$\frac{4}{7}\text{ sq.}\text{units}$

Answer

Correct option: B.

$\frac{7}{6}\text{ sq.}\text{units}$

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MCQ 1151 Mark

For which of the following values of mm, is the area of the region bounded by the curve $y = x - x^2$, and the line $y = mx$ equals $\frac{9}{2}\text{ sq. unit?}$

A
$-4$
B
$-2$
C
$2$
✓
$4$

Answer

Correct option: D.

$4$

The two curves meet at $mx = x - x^2$ or $= x^2 = x (1 - m)$
$\Rightarrow x^2 = x - mx$
$\therefore x = 0, 1 - m$
$=\int\limits^{1-\text{m}}_0(\text{y}_1-\text{y}_2)\text{dx}$
$=\Big[(1-\text{m)}\frac{\text{x}^2}{2}-\frac{\text{x}^2}{3}\Big]^{1-\text{m}}_0$
$=\frac{9}{2}(\text{given)}\text{ If}\text{ m}<1$
$=$ or $=(1-\text{m)}^3\Big[\frac{1}{2}-\frac{1}{3}\Big]=\frac{9}{2}$
$=$ or $-(1-\text{m)}^3=-27$
$=$ or $1-\text{m}=-3$
$\Rightarrow\text{m}=4$

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MCQ 1161 Mark

The area of ellipse $\frac{\text{x}^2}{4^2}+\frac{\text{y}^2}{9^2}=1$ is:

A
$6\pi\text{ sq}.\text{units}$
B
$\frac{\pi(\text{a}^2+\text{b}^2)}{4}\text{ sq}.\text{units}$
C
$\pi(\text{a+b})\text{ sq}.\text{units}$
✓
None of these

Answer

Correct option: D.

None of these

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MCQ 1171 Mark

Area between the curve $\text{y}=\cos^2\text{x},x-$axis and ordinates $x = 0$ and $x = p$ in the interval $(0, p)$ is:

A
$2\pi^3$
B
$2\pi$
C
$\pi$
✓
$\frac{\pi}{2}$

Answer

Correct option: D.

$\frac{\pi}{2}$

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MCQ 1181 Mark

The area included between the parabolas $y^2 = 4x$ and $x^2 = 4y$ is $($in square units$)$

A
$\frac{4}{3}$
B
$\frac{1}{3}$
✓
$\frac{16}{3}$
D
$\frac{8}{3}$

Answer

Correct option: C.

$\frac{16}{3}$

We have, $\text{x} = \frac{\text{y}^{2}}{4}\ ....(1)$
$\text{x}^{2} = 4\text{y}\ ....(2)$
points of intersection of two parabola is given by, $y^24^2 = 4y$
$\Big(\frac{\text{y}^{2}}{4}\Big)^{2} = 4\text{y}$
$\Rightarrow\text{y}^{4} -64\text{y} = 0$
$\Rightarrow\text{y}(\text{y}^{3}-64) = 0$
$\Rightarrow\text{y} = 0, 4$
$\Rightarrow\text{x} = 0, 4$
Therefore, the points of intersection are $A(0, 0)$ and $C(4, 4).$
Therefore, the area of the required region $\text{ABCD},$
$= \int\limits^4_0\Big(2\sqrt{\text{x}}-\frac{\text{x}^{2}}{4}\Big) \text{dx}$
$= \bigg[2\times\frac{2\text{x}^\frac{3}{2}}{3}-\frac{\text{x}^{3}}{12}\bigg]^4_0$
$=\bigg(2\times\frac{2(4)^\frac{3}{2}}{3}-\frac{(4)^{3}}{12}-\frac{(0)^{3}}{12}\bigg)$
$= \big(\frac{32}{3}-\frac{16}{3}\big)-0$
$= \frac{16}{3}\text{ square units}$

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MCQ 1191 Mark

The area bounded by the curve $y = 4x - x^2$ and the $x-$ axis is:

A
$\frac{30}{7}\text{ sq. units}$
B
$\frac{31}{7}\text{ sq. units}$
✓
$\frac{32}{3}\text{ sq. units}$
D
$\frac{34}{3}\text{ sq. units}$

Answer

Correct option: C.

$\frac{32}{3}\text{ sq. units}$

Point of intersection of parabola
$y = 4x - x^2$ with $x-$ axis is given by $y = 4x - x^2$ and $y = 0$
Equation of $x$ axis
$\Rightarrow 4x - x^2 = 0$
$\Rightarrow x = 0$ or $x = 4$
$\Rightarrow y = 0, y = 0$
Thus $0 (0, 0)$ and $B (4, 0)$ are points of intersection of parabola and $x -$ axis.
Required shaded area $= \int\limits^4_0(4\text{x}-\text{x}^2)\text{dx}$
$= \Big[2\text{ x}^2-\frac{\text{x}^3}{3}\Big]^4_0$
$= 2\times16-\frac{16}{3}-0$
$=\frac{96-64}{3}$
$=\frac{32}{3}\text{ sq. units}$

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MCQ 1201 Mark

The area of the circle $x^2+ y^2 = 16$ enterior to the parabola $y^2 = 6x$ is :

A
$\frac{4}{3}\big(4\pi-\sqrt{3}\big)$
B
$\frac{4}{3}\big(4\pi+\sqrt{3}\big)$
✓
$\frac{4}{3}\big(8\pi-\sqrt{3}\big)$
D
$\frac{4}{3}\big(8\pi+\sqrt{3}\big)$

Answer

Correct option: C.

$\frac{4}{3}\big(8\pi-\sqrt{3}\big)$

Points of intersection of the parabola and the circle is obtained by solving the simultaneous equation
$x^2 + y^2 = 16$ and $y^2 = 6x$
$\Rightarrow x^2 + 6x = 16$
$\Rightarrow x^2 = 6x -16 = 0$
$\Rightarrow (x + 8)(x - 2) = 0$
$⇒ x = 2$ or $x = -8$
$⇒ x = 2$ or $x = -8, x$ can not be $-8$ as in this case it will be the point outside circle.
$\therefore \text{x} = 2$
$\therefore \text{when } \text{x} = 2,\text{ y}=\pm\sqrt{6\times2}=\pm\sqrt{12}=\pm2\sqrt{3}$
$\therefore\text{B}(2, 2\sqrt{3})$ and $\text{ B}'(2,-2\sqrt{3})$ are points of intersection of the parabola and circle.
Required area
$=$ Area $\text{OB'C' A'CBO}$
$=$ area of circle $-$ area $\text{OBAB'O}$ Area of circle with radius $4$
$= \pi\times4^2$
$=16\pi$
Now,
Area $\text{OBAB'O}$
$= 2$ area $\text{OBAO}$
$= 2$ area $\text{OBDO} +$ area $\text{DBAD}$
$= 2\times\Bigg[\int\limits^2_0\sqrt{6\text{x}}\text{dx}+ \int\limits^4_2\sqrt{16-\text{x}^{2}}\Bigg]$
$= 2\times\Bigg\{\Bigg[\sqrt{6}\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}\Bigg]^2_0+ \Big[\frac{\text{x}}{2}\sqrt{16-\text{x}^2}+\frac{1}{2}\times16\sin^{-1}\Big(\frac{\text{x}}{4}\Big)\Big]^4_2\Bigg\}$
$= 2\times\Big\{\Big(\sqrt{6}\times\frac{2}{3}\times2^\frac{3}{2}-0\Big)+ \Big(\frac{1}{2}4\sqrt{16-(4)^2}\frac{1}{2}\times16\sin^{-1}\frac{4}{4}\\-\frac{2}{2}\sqrt{16-2^2}-\frac{1}{2}\times16\sin^{-1}\frac{2}{4}\Big)\Big\}$
$= 2\times\bigg[\Big(\sqrt{6}\times\frac{2}{3}\times2\sqrt{2}\Big)+0+8\sin^{-1}(1)-\sqrt{12}-8\sin^{-1}\Big(\frac{1}{2}\Big)\bigg]$
$= 2\times \bigg[\frac{8\sqrt{3}}{3}+8\times\frac{\pi}{2}-2\sqrt{3}-8\frac{\pi}{6}\bigg]$
$= 2\bigg\{\frac{8\sqrt{3}-6\sqrt{3}}{3}+8\big(\frac{\pi}{2}-\frac{\pi}{2}\big)\bigg\}$
$= 2\bigg\{\frac{2\sqrt{3}}{3}+8\Big(\frac{2\pi}{6}\Big)\bigg\}$
$= \frac{4\sqrt{3}}{3}+\frac{16\pi}{3}$

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MCQ 1211 Mark

If $A_n$ be the area bounded by the curve $y = (\tan x)^n$ and the lines $x = 0, y = 0$ and $\text{x}=\frac{\pi}{4},$ then for $x > 2$

✓
$\text{A}_{\text{n}}+\text{A}_{\text{n}-2}=\frac{1}{\text{n}-1}$
B
$\text{A}_{\text{n}}+\text{A}_{\text{n}-2}<\frac{1}{\text{n}-1}$
C
$\text{A}_{\text{n}}-\text{A}_{\text{n}-2}=\frac{1}{\text{n}-1}$
D
none of these

Answer

Correct option: A.

$\text{A}_{\text{n}}+\text{A}_{\text{n}-2}=\frac{1}{\text{n}-1}$

An $=$ Area bounded by the curve $\text{y}=\big\{\tan(\text{x})\big\}^\text{n}=\tan^\text{n}\text{(x)}$ and the lines $x = 0, y = 0,$ and $\text{x}=\frac{\pi}{4}.$
Therefore,
$\text{A}_\text{n}=\int\limits_\text{0}^{\frac{\pi}{4}}\tan^\text{n}\text{(x)}\text{dx}$
$\Rightarrow \text{A}_{\text{n}-2}=\int\limits_\text{0}^{\frac{\pi}{4}}\tan^{\text{n}-2}\text{(x)}\text{dx}$
Comsider, $\text{A}_\text{n}=\int\limits_\text{0}^{\frac{\pi}{4}}\tan^\text{n}\text{(x)}\text{dx}$
$\Rightarrow\ \text{A}_\text{n}=\int\limits_0^{\frac{\pi}{4}}\big\{\tan^{\text{n}-2}(\text{x})\big\}\big\{\tan^2(\text{x})\big\}\text{dx}$
$\Rightarrow\text{A}_\text{n}=\int\limits_0^\frac{\pi}{4}\big\{\tan^{\text{n}-2}(\text{x})\big\}\big\{\sec^2\text{(x)}-1\big\}\text{dx}$
$\Rightarrow\text{A}_\text{n}=\int\limits_0^\frac{\pi}{4}\big\{\tan^{\text{n}-2}\text{(x)} \sec^2(\text{x})-\tan^{\text{n}-2}(\text{x})\big\}\text{dx}$
$\Rightarrow\text{A}_\text{n}=\int\limits_0^{\frac{\pi}{4}}\big\{\tan^{\text{n}-2}\text{(x)}\sec^2(\text{x})\big\}\text{dx}-\int_0^\frac{\pi}{4}\tan^{\text{n}-2}\text{(x)}\text{ dx}$
$\Rightarrow\text{A}_\text{n}+\text{A}_{\text{n}-2}=\int\limits_0^\frac{\pi}{4}\tan^{\text{n}-2}(\text{x})\sec^2\text{(x) dx}$
Now, $\text{A}_\text{n}+\text{A}_{\text{n}+2}=\int\limits_0^\frac{\pi}{4}\tan^{\text{n}-2}\text{(x)}\sec^2(\text{x})\text{dx}$
Let $\text{u}=\tan\text{(x)}$
$\Rightarrow\text{du}=\sec^2\text{x }\text{dx}$
Also, when $x = 0, u = 0$ and when $\text{x}=\frac{\pi}{4},\text{u}=1$
Therefore,
$\text{A}_\text{n}+\text{A}_{\text{n}-2}=\int\limits_0^{\frac{\pi}{4}}\tan^{\text{n}-2}\text{(x)}\sec^2\text{(x) dx}$
$=\int\limits_0^1(\text{u}^{\text{n}-2})\text{du}$
$=\Big[\frac{\text{u}^{\text{n}-1}}{\text{n}-1}\Big]_0^1$
$=\Big[\frac{1}{\text{n}-1}-0\Big]=\frac{1}{\text{n}-1}$

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MCQ 1221 Mark

Area bounded between the parabola $y^2 = 4ax$ and its latus rectum is:

A
$\frac{1}{3}\text{a }\text{sq}.\text{units}$
B
$\frac{1}{3}\text{a}^2\text{ sq}.\text{units}$
C
$\frac{8}{3}\text{a}\text{ sq}.\text{units}$
✓
$\frac{8}{3}\text{a}^2\text{ sq}.\text{units}$

Answer

Correct option: D.

$\frac{8}{3}\text{a}^2\text{ sq}.\text{units}$

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MCQ 1231 Mark

Area bounded by the ellipse $\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}=1$ is.

✓
$6\pi\text{ sq.}\text{units}$
B
$3\pi\text{ sq.}\text{units}$
C
$12\pi\text{ sq.}\text{units}$
D
None of these

Answer

Correct option: A.

$6\pi\text{ sq.}\text{units}$

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MCQ 1241 Mark

The area $(sq.$ units$)$ bounded by the parabola $y^2 = 4ax$ and the line $x = a$ and $x = 4a$ is:

A
$\frac{\text{35a}^2}{3}$
B
$\frac{4\text{a}^2}{3}$
C
$\frac{7\text{a}^2}{3}$
✓
$\frac{56\text{a}^2}{3}$

Answer

Correct option: D.

$\frac{56\text{a}^2}{3}$

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MCQ 1251 Mark

Area bounded by the curve $\text{y}=\cos\text{x}$ between $\text{x}=0$ and $\text{x}=\frac{3\pi}{2}$ is:

A
$1 \ sq.$ unit
B
$2 \ sq.$ units
✓
$3 \ sq.$ units
D
$4 \ sq.$ units

Answer

Correct option: C.

$3 \ sq.$ units

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MCQ 1261 Mark

Consider the curves $\text{y}=\sin\text{x}$ and $\text{y}=\cos\text{x}.$ What is the area of the region bounded by the above two curves and the lines $\text{x}=0$ and $\text{x}=\frac{\pi}{4}?$

✓
$\sqrt{2}-1$
B
$\sqrt{2}+1$
C
$\sqrt{2}$
D
$2$

Answer

Correct option: A.

$\sqrt{2}-1$

The area enclosed by $\text{y}=\sin\text{x,}\text{ y}=\cos\text{x},\text{ x}=0,\text{x}=\frac{\pi}{4}$ is given by
$=\int\limits^\frac{\pi}{4}_0(\cos\text{x}-\sin\text{x})\text{dx}=\sin\text{x}+\cos\text{x}\Big|^\frac{\pi}{4}_0$

$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-=\sqrt{2}-1$

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MCQ 1271 Mark

Find the area of bounded by $\text{y}=\sin\text{x}$ from $\text{x}=\frac{\pi}{4}$ to $\text{x}=\frac{\pi}{2}:$

✓
$\frac{\sqrt{2-1}}{\sqrt2}$
B
$\frac{1}{2}$
C
$\frac{1}{4}$
D
$\text{none}\text{ of}\text{ these}$

Answer

Correct option: A.

$\frac{\sqrt{2-1}}{\sqrt2}$

The area bounded is given as
$=\int\limits^\frac{\pi}{2}_\frac{\pi}{4}\sin\text{ x}\text{dx}$
$=\cos\text{ x}\Big|^\frac{\pi}{2}_\frac{\pi}{4}$
$=\cos\frac{\pi}{2}+\cos\frac{\pi}{4}$
$=\frac{\sqrt{2}-1}{\sqrt{2}}$

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MCQ 1281 Mark

The area of the region bounded by the ellipse $\text{x}^\frac{2}{16}+\text{y}^\frac{2}{9}=1$ is:

✓
$12\pi$
B
$3\pi$
C
$24\pi$
D
$\pi$

Answer

Correct option: A.

$12\pi$

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MCQ 1291 Mark

Smaller area enclosed by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ is :

A
$2(\pi-2)$
✓
$\pi-2$
C
$\pi-1$
D
$2(\pi+2)$

Answer

Correct option: B.

$\pi-2$

We have, $x^2 + y^2 = 4$ represents a circle with centre at $O(0, 0)$ and radius $2$
$x + y = 2$ represents a straight line cutting the $x-$ axis at $A(2, 0)$ and $y$ axis at $B(0, 2)$
Thus, $A(2, 0)$ and $B(0, 2)$ are also the points of intersection of the straight line and the circle smaller area enclosed by the by the curve and straight line is the shaded area.
shaded area $\text{(ABCA)} =$ area $\text{(OBCA)} - $ area $\text{(OBAO)}$
$=\int\limits^2_0\sqrt{4-\text{x}^2}\text{dx}-\int\limits^2_0(2 - \text{x})\text{dx}$
$\big[\therefore\text{x}^2 + \text{y}^2=4$
$\Rightarrow\text{y}=\sqrt{4-\text{x}^2}\text{ and }\text{x}+\text{y} = 2$
$\Rightarrow\text{y} = 2-\text{x}\big]$
$=\int\limits^2_0\Big[\big(\sqrt{4-\text{x}^2}\big)+\text{x} - 2\Big]$
$= \bigg[\frac{1}{2}\text{x}\sqrt{4-\text{x}^2}+\frac{1}{2}\times4\times\sin^{-1}\Big(\frac{\pi}{2}\Big)+\Big(\frac{\text{x}^2}{2}-2\text{x}\Big)\bigg]^2_0$
$= \frac{1}{2}\times2\sqrt{4-2^2}+2\times\sin^{-1}\Big(\frac{2}{2}\Big)+\Big(\frac{2^2}{2}-2\times2\Big)-0$
$= 0 + 2 \times\frac{\pi}{2}+(2 - 4)$
$= (\pi-2) \text{sq. units}$

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MCQ 1301 Mark

A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{32}=1$ ntersects the major and minor axes at points $A$ and $B$ respectively. If $C$ is the center of the ellipse, then area of the triangle $\text{ABC}$ is:

A
$12$ sq. units
✓
$24$ sq. units
C
$36$ sq. units
D
$48$ sq. units

Answer

Correct option: B.

$24$ sq. units

One of the tangents of slope $m$ to the given ellipse is
$=\text{y}=\text{mx}+\sqrt{18\text{m}^2+32}\text{ for}$
$=\text{m}=-\frac{4}{3},\text{we}\text{ have}$
$=\text{y}=-\frac{4}{3}\text{x}+8.$
Then points on the axis where tangents meet are $A(6, 0)$ and $B(0, 8).$
Then area of triangle $\text{ABC}$ is
$=\frac{1}{2}(6)(8)=24\text{ units}.$

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MCQ 1311 Mark

Choose the correct answer from the given four options : Area of the region in the first quadrant enclosed by the $x-$ axis, the line $y = x $ and the circle $x^2 + y^2 = 32$ is:

A
$16\pi\text{ sq. units}$
✓
$4\pi\text{ sq. units}$
C
$32\pi\text{ sq. units}$
D
$24\pi\text{ sq. units}$

Answer

Correct option: B.

$4\pi\text{ sq. units}$

We have, $y = 0, y = x$ and the circle $x^2 + y^2 - 3$ in the first quadrant

Solving $y = x$ with the circle
$x^2 + x^2 = 32$
$\Rightarrow x^2 = 16$
$\Rightarrow x = 4 ($In first quadrant$)$
When $x = 4, y = 4$
For point of intersection of circle with the $x-$ axis,
Put $y = 0$
$\therefore\ \text{x}^2+0=32$
$\Rightarrow\ \text{x}=\pm4\sqrt{2}$
So, the circle intersects the $x-$ axis at $\big(\pm4\sqrt{2},0\big)$
From the figure, area of shaded region
$\text{A}=\int\limits^4_0\text{x dx}+\int\limits^{4\sqrt{2}}_4\sqrt{(4\sqrt{2})^2-\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}^2}{2}\Big]^4_0+\bigg[\frac{\text{x}}{2}\sqrt{(4\sqrt{2})^2-\text{x}^2}+\frac{\big(4\sqrt{2}\big)^2}{2}\sin^{-1}\frac{\text{x}}{4\sqrt{2}}\bigg]^{4\sqrt{2}}_0$
$=\frac{16}{2}+\bigg[0+16\sin^{-1}1-\frac{4}{2}\sqrt{\big(4\sqrt{2}\big)^2-16}-16\sin^{-1}\frac{4}{4\sqrt{2}}\bigg]$
$=8+\bigg[16\cdot\frac{\pi}{2}-2\cdot\sqrt{16}-16\cdot\frac{\pi}{4}\bigg]$
$=8+\big[8\pi-8-4\pi\big]=4\pi\text{ sq. units}$

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MCQ 1321 Mark

Area of triangle whose two vertices formed from the $x-$axis and line $y = 3 - |x|$ is:

A
$\text{9 sq. units}$
B
$\frac{3}{2}\text{sq}.\text{units}$
C
$\text{3 sq. units}$
✓
None of these

Answer

Correct option: D.

None of these

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MCQ 1331 Mark

The area enclosed by the curve $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$ is:

A
$10\pi\text{ sq.}\text{units}$
✓
$15\pi\text{ sq.}\text{units}$
C
$5\pi\text{ sq.}\text{units}$
D
$4\pi\text{ sq.}\text{units}$

Answer

Correct option: B.

$15\pi\text{ sq.}\text{units}$

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MCQ 1341 Mark

The area bounded by the $x -$ axis, the curve $y = f(x)$ and the lines $x = 1, x = b$ is equal to $\sqrt{\text{b}}^2+1-\sqrt{2}$ for all $ b>,$then$\text{ f(x)}\text{ is:}$

A
$\sqrt{\text{x}-1}$
B
$\sqrt{\text{x}+1}$
C
$\sqrt{\text{x}^2+1}$
✓
$\text{x}\sqrt{\text{x}-1}$

Answer

Correct option: D.

$\text{x}\sqrt{\text{x}-1}$

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MCQ 1351 Mark

The area bounded by $= 4ax$ and $y = mx$ is $\frac{\text{a}^2}{3}\text{sq. units}$ then $m:$

A
$1$
✓
$2$
C
$3$
D
$4$

Answer

Correct option: B.

$2$

The two curves $y^2 = 4ax$ and $y = mx$ intersect at
$=\Big(\frac{4\text{a}}{\text{m}^2},\frac{4\text{a}}{\text{m}}\Big)$ and the area enclosed by the two curves are given by
$=\int\limits^\frac{4\text{a}}{\text{m}2}_0\Big(\sqrt{4\text{ax}}-\text{mx}\Big)\text{dx}$
$\therefore\int\limits^\frac{4\text{a}}{\text{m}^2}\Big(\sqrt{4\text{ax}}-\text{mx}\Big)\text{dx}=\frac{\text{a}^2}{3}$
$\Rightarrow\frac{8}{3}.\frac{\text{a}^2}{\text{m}^3}=\frac{\text{a}^2}{3}$
$\Rightarrow\text{m}^3=8$
$\Rightarrow\text{m}=2$

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MCQ 1361 Mark

Area lying between the parabola $y^2 = 4x$ and its latus rectum is:

A
$\frac{1}{3}\text{ sq.}\text{units}$
B
$\frac{2}{3}\text{ sq.}\text{units}$
C
$\frac{5}{3}\text{ sq.}\text{units}$
✓
$\frac{8}{3}\text{ sq.}\text{units}$

Answer

Correct option: D.

$\frac{8}{3}\text{ sq.}\text{units}$

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MCQ 1371 Mark

If area bounded by the curves $x = at^2$ and $y = ax^2$ is $1,$ then a $.......$

A
$\frac{1}{2}$
B
$\frac{1}{3}$
✓
$\frac{1}{\sqrt{3}}$
D
$1$

Answer

Correct option: C.

$\frac{1}{\sqrt{3}}$

$\text{x}=\text{ay}^2$ and ${ y}=\text{ax}^2$
$\Rightarrow\text{a}^3\text{ x}^3=1$
$\Rightarrow\text{(x},\text{y)}=\Big(\frac{1}{\text{a}},\frac{1}{\text{a}}\Big)$
The area bounded by the curves is computed by:
$=\int\limits^\frac{1}{\text{a}}_0\text{a}\text{x}^2-\sqrt{\frac{\text{x}}{\text{a}}}\text{ dx}=1$
$\Rightarrow\frac{1}{3\text{a}^2}=1$
$\Rightarrow\text{a}=\frac{1}{\sqrt{3}}$

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MCQ 1381 Mark

Area of curve explained in the passage from $0$ to $\frac{\pi}{2}\text{ is:}$

A
$\frac{1}{3}\text{ sq.}\text{ unit}$
B
$\frac{1}{2}\text{ sq.}\text{ unit}$
✓
$1\text{ sq.}\text{ unit}$
D
$2\text{ sq.}\text{ units}$

Answer

Correct option: C.

$1\text{ sq.}\text{ unit}$

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MCQ 1391 Mark

The area bounded by the curve $\text{y}=\sec^2\text{x},\text{y}$ and $\text{x}=\frac{\pi}{3}$ is:

A
$\sqrt{3}\text{ sq.}\text{ units}$
B
$\sqrt{2}\text{ sq.}\text{ units}$
✓
$2\sqrt{3}\text{ sq.}\text{ units}$
D
None of these

Answer

Correct option: C.

$2\sqrt{3}\text{ sq.}\text{ units}$

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MCQ 1401 Mark

The area bounded by the curves $x + 2y^2 = 0$ and $x + 3y^2 = 1$ is:

A
$1\text{ sq.}\text{units}$
B
$\frac{1}{3}\text{ sq.}\text{units}$
C
$\frac{2}{3}\text{ sq.}\text{units}$
✓
$\frac{4}{3}\text{ sq.}\text{units}$

Answer

Correct option: D.

$\frac{4}{3}\text{ sq.}\text{units}$

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MCQ 1411 Mark

The area of the region $($in square units$)$ bounded by the curve $x^2 = 4y$, line $x = 2$ and $x-$axis is:

A
$1$
✓
$\frac{2}{3}$
C
$\frac{4}{3}$
D
$\frac{8}{3}$

Answer

Correct option: B.

$\frac{2}{3}$

$x^2 = 4y$ and $x= 2$
$\Rightarrow 4 = 4y$
$\Rightarrow y = 1$
$A(2, 1)$ is the point of intersection of curve and straight the
Area of shaded region $\text{OAB} = \int\limits^2_0\text{y}\text{ dx}$
$=\int\limits^2_0\frac{\text{x}^2}{4}\text{ dx}$
$=\Big[\frac{\text{x}^3}{12}\Big]^2_0$
$= \frac{2^3}{12}-0$
$= \frac{2}{3}\text{ square units}$

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MCQ 1421 Mark

Choose the correct answer from the given four options:
The area of the region bounded by the curve $\text{y}=\sqrt{16-\text{x}^2}$ and x-axis is:

✓
$8\text{ sq. units}$
B
$20\pi\text{ sq. units}$
C
$16\pi\text{ sq. units}$
D
$256\pi\text{ sq. units}$

Answer

Correct option: A.

$8\text{ sq. units}$

Given equation of curve is $\text{y}=\sqrt{16-\text{x}^2}$ and the equation of line is

X-axis i.e., y = 0

$\therefore\ \sqrt{16-\text{x}^2}=0\ \ \dots(\text{i})$

$\Rightarrow\ 16-\text{x}^2=0$

$\Rightarrow\ \text{x}^2=16$

$\Rightarrow\ \text{x}=\pm4$

So, the intersection points are (4, 0) and (-4, 0).

$\therefore$ Area of curve, $\text{A}=\int\limits^4_{-4}(16-\text{x}^2)^{\frac{1}{2}}\text{dx}$

$=\int\limits^4_{-4}\sqrt{(4^2-\text{x}^2)}\text{dx}$

$=\bigg[\frac{\text{x}}{2}\sqrt{4^2-\text{x}^2}+\frac{4^2}{2}\sin^{-1}\frac{\text{x}}{4}\bigg]^4_{-1}$

$=\bigg[\frac{4}{2}\sqrt{4^2-4^2}+8\sin^{-1}\frac{4}{4}\bigg]-\bigg[-\frac{4}{2}\sqrt{4^2-(-4)^2}+8\sin^{-1}\Big(-\frac{4}{4}\Big)\bigg]$

$=\bigg[2\cdot0+8\cdot\frac{\pi}{2}-0+8\cdot\frac{\pi}{2}\bigg]=8\pi\text{ sq. units}$

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MCQ 1431 Mark

Area between the curves $y = x$ and $y = x^3$ is:

✓
$\sqrt{3}\sqrt{2}$
B
$\frac{1}{2}$
C
$\frac{2}{\sqrt{2}}$
D
$\frac{1}{4}$

Answer

Correct option: A.

$\sqrt{3}\sqrt{2}$

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MCQ 1441 Mark

The area of region bounded by curve $\text{y}=\cos2\text{x},$ line $\text{x}=0$ $\text{x}=\frac{\pi}{3}$ is:

A
$\frac{2-\sqrt{3}}{4}$
✓
$\frac{\sqrt{3}}{4}$
C
$\frac{4-\sqrt{3}}{4}$
D
$\frac{\sqrt{3}-4}{4}$

Answer

Correct option: B.

$\frac{\sqrt{3}}{4}$

$=\text{y}=\cos2\text{x}\int\limits^\frac{\pi}{3}_0\cos2\text{ xdx}$

$=\frac{1}{2}[\sin2\text{x}]^\frac{\pi}{3}_0$

$=\frac{\sqrt{3}}{4}\text{sq}\text{ units}$

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MCQ 1451 Mark

Area of the region bounded by the curve $\text{y}=\tan\text{x,}$ line $\text{x}=\frac{\pi}{4}$ and the $x-$axis is:

A
$\log2\text{ sq.}\text{units}$
✓
$\frac{1}{2}\log2\text{ sq.}\text{units}$
C
$\frac{1}{3}\log2\text{ sq.}\text{units}$
D
$5\log2\text{ sq.}\text{units}$

Answer

Correct option: B.

$\frac{1}{2}\log2\text{ sq.}\text{units}$

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MCQ 1461 Mark

The area of the region bounded by the curve $x = 2y + 3$ and lines $y = 1$ and $y = –1$ is:

A
$4\text{ sq.}\text{ units}$
B
$\frac{2}{3}\text{ sq.}\text{ units}$
✓
$6\text{ sq.}\text{ units}$
D
$8\text{ sq.}\text{ units}$

Answer

Correct option: C.

$6\text{ sq.}\text{ units}$

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MCQ 1471 Mark

The area bounded by the curve y = f(x), above the x - axis, between ax = a and x = b is:

A
$\int\limits^{\text{b}}_{\text{f(a)}}\text{ydy}$
B
$\int\limits^{\text{fb}}_{\text{(b)}}\text{xdx}$
C
$\int\limits^{\text{b}}_{\text{a}}\text{xdy}$
✓
$\int\limits^{\text{b}}_{\text{a}}\text{ydx}$

Answer

Correct option: D.

$\int\limits^{\text{b}}_{\text{a}}\text{ydx}$

We need to calculate the area bounded by y = f(x) between x = a and x = bx = b
So, we need to find the area of the region inside the curve y = f(x)
having limit points aa to bb and it is given by
$\int\limits^\text{b}_\text{a}=\text{f(x)}\text{ dx}=\int\limits^\text{b}_\text{a}$

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MCQ 1481 Mark

Area enclosed between the curve $y^2(2a - x) = x^3$ and the line $x = 2a$ above $x-$ axis is:

A
$\pi\text{a}^2$
✓
$\frac{3}{2}\pi\text{a}^2$
C
$2\pi\text{a}^2$
D
$3\pi\text{a}^2$

Answer

Correct option: B.

$\frac{3}{2}\pi\text{a}^2$

$\text{y}^2 (2\text{a}-\text{x})=\text{x}^3$
$\text{y}= \sqrt\frac{\text{x}^3}{2\text{a}-\text{x}}$
$\text{Let}\text{ x}= 2\text{a}\sin^2\theta$
$\text{dx}=4\text{a}\sin\theta\cos\theta\text{d}\theta$
$\text{Area}=\int\limits^\text{2a}_0\sqrt\frac{\text{x}^3}{2\text{a}-\text{x}}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\sqrt\frac{(8\text{a}^3)\sin^6\theta}{(2\text{a})\cos^2\theta}.(4\text{a})\sin\theta\cos\theta\text{d}\theta$
$= 8\text{a}^2\int^\frac{\pi}{2}_0\sqrt{\sin^6}\theta\sin\theta\text{d}\theta$
$=8\text{a}^2\Big[\int^\frac{\pi}{2}_0\sin^4\theta\text{d}\theta\Big]$
$= 8\text{a}^2\Big[\int^\frac{\pi}{2}_0\sin^2\theta(1-\cos^2\theta)\text{d}\theta\Big]$
$= 8\text{a}^2\Big[\int^\frac{\pi}{2}_0\frac{(1-\cos2\theta)}{2}\text{d}\theta-\frac{1}{4}\int^\frac{\pi}{2}_0\sin^2\theta\text{d}\theta\Big]$
$= 8\text{a}^2\bigg[\frac{1}{2}[\theta]^\frac{\pi}{2}_0-\Big[\frac{\sin2\theta}{4}\Big]^\frac{\pi}{2}_0\bigg]-\frac{1}{4}\bigg[\int^\frac{\pi}{2}_0\frac{1-\cos4\theta}{2}\text{d}\theta\bigg]$
$= 8\text{a}^2\Big[\Big(\frac{\pi}{4}\Big)-0\Big]-\frac{1}{4}\Big[\frac{\pi}{4}-0\Big]$
$= 8\text{a}^2\Big[\frac{\pi}{4}-\frac{\pi}{16}\Big]$
$=\frac{3}{2}\pi\text{a}^2$

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MCQ 1491 Mark

The area of the region bounded by the curves $y =| x – 2 |, x = 1, x = 3$ and the $x-$axis is:

A
$4$
B
$2$
C
$3$
✓
$1$

Answer

Correct option: D.

$1$

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MCQ 1501 Mark

Area bounded by the ellipse $\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}=1$

✓
$6\pi\text{ sq}.\text{units}$
B
$3\pi\text{ sq}.\text{units}$
C
$12\pi\text{ sq}.\text{units}$
D
None of these

Answer

Correct option: A.

$6\pi\text{ sq}.\text{units}$

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MCQ 1511 Mark

Area of the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ is:

A
$4\pi\text{ ab}\text{ sq}.\text{units} $
B
$2\pi\text{ ab}\text{ sq}.\text{units} $
✓
$\pi\text{ ab}\text{ sq}.\text{units} $
D
$\frac{\pi\text{ab}}{2}\text{ sq}.\text{units}$

Answer

Correct option: C.

$\pi\text{ ab}\text{ sq}.\text{units} $

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MCQ 1521 Mark

The area bounded by the curve $y = x^2 - 1$ and the straight line $x + y = 3$ is:

A
$\frac{9}{2}\text{sq}.\text{units}$
B
$4\text{ sq}.\text{units}$
C
$\frac{9\sqrt{17}}{6}\text{sq}.\text{units}$
✓
$\frac{17\sqrt{17}}{6}\text{sq}.\text{units}$

Answer

Correct option: D.

$\frac{17\sqrt{17}}{6}\text{sq}.\text{units}$

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MCQ 1531 Mark

Area of the region between the curves $\text{x}^2+\text{y}^2=\pi,\text{y}=\sin\text{x}$ and $y-$axis in first quadrant is:

✓
$\frac{\pi^3-8}{4\text{ sq.}\text{ units}}$
B
$\frac{\pi^3-4}{4\text{ sq.}\text{ units}}$
C
$\frac{\pi^3-8}{4\text{ sq.}\text{ units}}$
D
$\frac{\pi^3-4}{4\text{ sq.}\text{ units}}$

Answer

Correct option: A.

$\frac{\pi^3-8}{4\text{ sq.}\text{ units}}$

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MCQ 1541 Mark

The area bounded by the curve $y = x|x|$ and the ordinates $x = -1$ and $x = 1$ is given by:

A
$0$
B
$\frac{1}{3}$
✓
$\frac{2}{3}$
D
$\frac{4}{3}$

Answer

Correct option: C.

$\frac{2}{3}$

The given equation of the curve is
$y = x|x|$
$\Rightarrow \text{y}= \text{cases} \ \text{x}^2 \text{x} \geq0$
$\text{-x}^2 \text{x} < 0\ \text{cases}$
Now, solving $x = 1$ and $y = x|x|$ we get
$x = 1 \Rightarrow y = 1$
$\Rightarrow A(1, 1)$ is point of intersection of the curve $y = x|x|$ and $x = 1$
Also, solving $x = -1$ and $y = x|x|$ we get
$x = -1$
$ \Rightarrow y = -1$
$\Rightarrow A'(-1, -1)$ is point of intersection of the curve $y = x |x|$ and $x = -1$
If $P(x, y_1), x > 0$ is a point on $y = x|x|$ then $y_1 >0 \Rightarrow |y_1| = y_1$_
And $Q(x, y_2), x < 0$ is a point on $y = x|x|$ then $y_2 < 0 \Rightarrow |y_2| = -y_2$
Required area $=\int\limits^0_{-1}|\text{y}_2|\text{dx}+\int\limits^1_0|\text{y}_1|\text{dx}$
$= \int\limits^0_{-1}-\text{y}_2\text{ dx}+\int\limits^1_0\text{y}_1\text{ dx}$
$= \int\limits^0_{-1}-(-\text{x}^{2})\text{dx}+\int\limits^1_0\text{x}^2\text{ dx}$
$=\int\limits^0_{-1}\text{x}^2\text{dx}+\int\limits^1_0\text{x}^2\text{dx}$
$= \Big[\frac{\text{x}^3}{3}\Big]^0_{-1} +\Big[\frac{\text{x}^3}{3}\Big]^1_0$
$= \bigg[0-\frac{(-1)^3}{3}\bigg]+\Big(\frac{1^3}{3}-0\Big)$
$= \frac{1}{3}+\frac{1}{3}$
$= \frac{2}{3}\text{ sq. units}$

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MCQ 1551 Mark

Area bounded by the curve $\text{y}=\log\text{x}$ and the coordinate axes is:

A
$2$
✓
$1$
C
$5$
D
$2\sqrt{2}$

Answer

Correct option: B.

$1$

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MCQ 1561 Mark

Choose the correct answer from the given four options:
The area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is:

✓
$\frac{7}{2}\text{ sq. units}$
B
$\frac{9}{2}\text{ sq. units}$
C
$\frac{11}{2}\text{ sq. units}$
D
$\frac{13}{2}\text{ sq. units}$

Answer

Correct option: A.

$\frac{7}{2}\text{ sq. units}$

From the figure, are of the shaded region,

$\text{A}=\int\limits^{3}_{2}(\text{x}+1)\text{dx}=\bigg[\frac{\text{x}^2}{2}+\text{x}\bigg]^3_2$ $=\bigg[\frac{9}{2}+3-\frac{4}{2}-2\bigg]=\frac{7}{2}\text{sq. units}$

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MCQ 1571 Mark

The area bounded by the curve $y = x^2 + 4x + 5,$ the axes of coordinates and minimum ordinate is:

A
$3\frac{2}{3}\text{sq}.\text{ units}$
✓
$4\frac{2}{3}\text{sq}.\text{ units}$
C
$5\frac{2}{3}\text{sq}.\text{ units}$
D
none of these

Answer

Correct option: B.

$4\frac{2}{3}\text{sq}.\text{ units}$

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MCQ 1581 Mark

The area of the region bounded by the curve $\text{y}=\sqrt{16-\text{x}^2}$ and $\text{ x}-$axis is:

✓
$\text{8p sq. units}$
B
$\text{20p sq. units}$
C
$\text{16p sq. units}$
D
$\text{256p sq. units}$

Answer

Correct option: A.

$\text{8p sq. units}$

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MCQ 1591 Mark

The area of the smaller region bounded by the ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=1$ and the line $\frac{\text{x}}{3}+\frac{\text{y}}{2}=1$ is:

✓
$3(\pi-2)$
B
$\frac{3}{2\pi}$
C
$\frac{3}{2}(\pi-2)$
D
$\frac{2}{3}(\pi-2)$

Answer

Correct option: A.

$3(\pi-2)$

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MCQ 1601 Mark

The area of the region bounded by the curve $x = 2y + 3$ and the lines $y = 1$ and $y = -1$ is:

A
$\text{4 sq. units}$
B
$\frac{3}{2}\text{sq.}\text{ units}$
✓
$\text{6 sq. units}$
D
$\text{8 sq. units}$

Answer

Correct option: C.

$\text{6 sq. units}$

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MCQ 1611 Mark

The area bounded by the y-axis, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x}$ when $0\leq\text{x}\leq\frac{\pi}{2}$ is:

A
$2\big(\sqrt{2}-1\big)$
✓
$\sqrt{2}-1$
C
$\sqrt{2}+1$
D
$\sqrt{2}$

Answer

Correct option: B.

$\sqrt{2}-1$

Points of intersection is obtained by solving y = sinx and y = cos x

$\therefore\sin\text{x} = \cos \text{x}$

$\Rightarrow \text{x}=\frac{\pi}{4}$

Thus the two functions intesect at $\text{x}=\frac{\pi}{4}$

$\Rightarrow \text{y} = \sin \frac{\pi}{4} =\frac{1}{\sqrt{2}}$

Hence $\text{A}\Big(\frac{\pi}{4},\frac{1}{\sqrt{2}}\Big)$ is the point of intersection.

$\therefore$ Area bound by the curves and the y - axis when $0\leq\text{x}\leq\pi2$

$\text{A} = \int\limits^\frac{1}{\sqrt{2}}_0|\text{x}_1|\text{dy}+\int\limits^1_\frac{1}{\sqrt{2}}|\text{x}_2|\text{dy}$

$=\int\limits^\frac{1}{\sqrt{2}}_0\text{x}_1\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\text{x}_2\text{ dy}$

$= \int\limits^\frac{1}{\sqrt{2}}_0\sin^{-1}\text{y}\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\cos^{-1}\text{y}\text{ dy}$

$= \Big[\text{y}\sin^{-1}\text{y}+\sqrt{1-\text{y}^2}\Big]^\frac{1}{\sqrt{2}}_0+\Big[\text{y}\cos^{-1}\text{y}-\sqrt{1-\text{y}^2}\Big]^1_\frac{1}{\sqrt{2}}$

$= \Big[\frac{1}{\sqrt{2}}\sin^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}-1\Big]\\+\bigg[1\times\cos^{-1}-0-\frac{1}{\sqrt{2}}\cos^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}\bigg]$

$= \frac{2}{\sqrt{2}}-1$

$=\big(\sqrt{2}-1\big)\text{sq. units}$

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MCQ 1621 Mark

Choose the correct answer in the following:
The area bounded by the y-axis, y = cos x and y = sin x when $0\leq\text{x}\leq\frac{\pi}{2}$

A
$2(\sqrt2-1)$
✓
$\sqrt2-1$
C
$\sqrt2+1$
D
$\sqrt2.$

Answer

Correct option: B.

$\sqrt2-1$

The given equations are
y = cos x ...(1)
And, y = sin x ...(2)

Required area = Area(ABLA) + Area(OBLO)]
$=\int\limits^1_{\frac{1}{\sqrt2}}\text{x dy}+\int\limits^{\frac{1}{\sqrt2}}_0\text{x dy}$
$=\int\limits^1_{\frac{1}{\sqrt2}}\cos^{-1}\text{y dy}+=\int\limits^{\frac{1}{\sqrt2}}_0\sin^{-1}\text{x dy}$
Integrating by parts, we obtain
$=\Big[\text{y}\cos^{-1}\text{y}-\sqrt{1-\text{y}^2}\Big]^1_{\frac{1}{\sqrt2}}$ $+\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1+\text{x}^2}\Big]^{\frac{1}{\sqrt2}}_0$
$=\Big[\cos^{-1}(1)-\frac{1}{\sqrt2}\cos^{-1}\Big(\frac{1}{\sqrt2}\Big)+\sqrt{1-\frac{1}{2}}\ \Big]$ $+\Big[\frac{1}{\sqrt2}\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)+\sqrt{1-\frac12}-1\Big]$
$=\frac{-\pi}{4\sqrt2}+\frac{1}{\sqrt2}+\frac{\pi}{4\sqrt2}+\frac{1}{\sqrt2}-1$
$=\frac{2}{\sqrt2}-1$
$=\sqrt2-1\text{ units}$
Thus, the correct answer is B.

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MCQ 1631 Mark

The area bounded by the circles $= 4x^2 + y^2 = 1, x^2 + y^2 = 4$ in the first Quadrant is:

A
$\frac{\pi}{2}$
✓
$\frac{3\pi}{4}$
C
$3\pi$
D
$\frac{\pi}{4}$

Answer

Correct option: B.

$\frac{3\pi}{4}$

$x^2+ y^2= 1$ and $x^2 + y^2 = 4$ are concentric circles They form Ring in between them.Area is given by
$=\pi(4-1)=3\pi$
In first Quadrant it is divided by $4$ So, Area is given by
$=\frac{3\pi}{4}$

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MCQ 1641 Mark

Let $f(x) = x^2 - 3x + 2$ then area bounded by the curve $f( ∣x∣ ) ($in square units$)$ and $x -$ axis is:

A
$\frac{1}{3}$
B
$\frac{5}{6}$
✓
$\frac{5}{3}$
D
none of these

Answer

Correct option: C.

$\frac{5}{3}$

the area bounded by the curve $y = f(x), x -$ axis and $y - $axis be a square units
then area bounded by $f(∣x∣)$ and $x -$ axis be twice of a Shaded Area
$=\int\limits^1_0\text {f(x)}\text{ dx}=\frac{5}{6}\text{sq}.$
units again Graph of $f(∣x∣)$
$\therefore$ Required Area bounded by $f(∣x∣)$
$=\int\limits^1_{-1}\text{f(dx)}=2$
$=\int\limits^1_0\text{f(dx)}=2,\frac{5}{6}$
$=\frac{5}{3}\text{sq}\text{ units}$

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MCQ 1651 Mark

The area of the portion of the circle $x^2 + y^2 = 1$, which lies inside the parabola $y^2 = 1 - x$, is:

A
$\frac{\pi}{2}-\frac{2}{3}$
B
$\frac{\pi}{2}+\frac{2}{3}$
✓
$\frac{\pi}{2}-\frac{4}{3}$
D
$\frac{\pi}{2}+\frac{4}{3}$

Answer

Correct option: C.

$\frac{\pi}{2}-\frac{4}{3}$

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MCQ 1661 Mark

Area of the region bounded by rays |x| + y = 1 and X - axis is ___________.

✓
$\frac{1}{2}$
B
$2$
C
$1$
D
$\frac{1}{4}$

Answer

Correct option: A.

$\frac{1}{2}$

Given equation can also be written as y = 1 - |x|
point of intersection of the given line are (1, 0) and (0, 1) and (0, 0) will be the third vertices of the triangle formed
thus height and base of this triangle = 1
therefore Area $=1\times1\times\frac{1}{2}$

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MCQ 1671 Mark

The area of the region bounded by the parabola $y = x^2 + 1$ and the straight line $x + y = 3$ is given by:

A
$\frac{45}{7}\text{sq.}\text{units}$
B
$\frac{25}{4}\text{sq.}\text{units}$
C
$\frac{5}{18}\text{sq.}\text{units}$
✓
$\frac{9}{2}\text{sq.}\text{units}$

Answer

Correct option: D.

$\frac{9}{2}\text{sq.}\text{units}$

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MCQ 1681 Mark

Choose the correct answer from the given four options:
The area of the region bounded by the ellipse $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{16}=1$ is:

✓
$20\pi\text{ sq. units}$
B
$20\pi^2\text{ sq. units}$
C
$16\pi^2\text{ sq. units}$
D
$25\pi\text{ sq. units}$

Answer

Correct option: A.

$20\pi\text{ sq. units}$

We have $\frac{\text{x}^2}{5^2}+\frac{\text{y}^2}{4^2}=1,$ which is ellipse with is axis as coordinate axis.

$\frac{\text{y}^2}{4^2}=1-\frac{\text{x}^2}{5^2}$

$\Rightarrow\ \text{y}^2=16\Big(1-\frac{\text{x}^2}{25}\Big) $

$\Rightarrow\ \text{y}=\frac{4}{5}\sqrt{5^2-\text{x}^2}$

From the figure, area of the shaded region

$\text{A}=4\int\limits_{0}^{5}\frac{4}{5}\sqrt{5^2-\text{x}^2}\text{ dx}$

$=\frac{16}{5}\bigg[\frac{\text{x}}{2}\sqrt{5^2-\text{x}^2}+\frac{5^2}{2}\sin^{-1}\frac{\text{x}}{5}\bigg]^{5}_{0} $

$=\frac{16}{5}\bigg[0+\frac{5^2}{2}\sin^{-1}1-0-0\bigg]$ $=\frac{16}{5}.\frac{25}{2}.\frac{\pi}{2}=20\pi\text{ sq. units}$

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MCQ 1691 Mark

Area of the region bounded by the curve x = 2y + 3, the y-axis and between y = -1 and y = 1 is:

A
$4\text{sq}\text{ units}3$
B
$\frac{3}{2}\text{sq}\text{ units}$
✓
$6\text{sq}\text{ units}$
D
$8\text{sq}\text{ units}$

Answer

Correct option: C.

$6\text{sq}\text{ units}$

$\text{as}\text{ area}=\int\limits^1_0(2\text{y}+3)\text{dy}$

$=6\text{sq}\text{ units}$

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MCQ 1701 Mark

The area of the region bounded by the line $y = | x - 2 |, x = 1, x = 3$ and $x-$axis is:

A
$\text{4 sq. units}$
B
$\text{2 sq. units}$
C
$\text{3 sq. units}$
✓
$\text{1 sq. unit}$

Answer

Correct option: D.

$\text{1 sq. unit}$

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MCQ 1711 Mark

The area bounded by the curve $\text{y}=\frac{3}{2}\sqrt{\text{x}},$ the line $\text{x}=1$ and $x -$ axis is $..........sq.$ units:

A
$2$
✓
$4$
C
$6$
D
$8$

Answer

Correct option: B.

$4$

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MCQ 1721 Mark

Area of ellips $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ is:

A
$4\pi\text{ ab}\text{ sq.}\text{ units}$
B
$2\pi\text{ ab}\text{ sq.}\text{ units}$
✓
$\pi\text{ ab}\text{ sq.}\text{ units}$
D
$\frac{\pi\text{ab}}{2}\text{ sq.}\text{ units}$

Answer

Correct option: C.

$\pi\text{ ab}\text{ sq.}\text{ units}$

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MCQ 1731 Mark

The area $($in $sq.$ units$)$ enclosed between the graph of $y = x^3$ and the lines $x = 0, y = 1, y = 8$ is:

✓
$\frac{45}{4}$
B
$14$
C
$7$
D
None of these

Answer

Correct option: A.

$\frac{45}{4}$

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MCQ 1741 Mark

What is the area of the triangle bounded by the lines $y = 0, x + y = 0$ and $x = 4?$

A
$4$ units
✓
$8$ units
C
$12$ units
D
$16$ units

Answer

Correct option: B.

$8$ units

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Maths M.C.Q (1 Marks)

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