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
View full question & 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
View full question & answer→Correct option: C.
$\frac { 9 }{ 2 }$
To find the Area of the region bounded by the curve $x=y^2-2$ and $x=y$ is :
First , we have to find the point of Intersection of the $2$ curves $y = y^2 - 2$
$y^2-y-2=0$
$y=-1, y=2$
Now , Area of the region bounded by the curve is
$=\int\limits^2_\text{y}=-1\Big[\text{y}-\text{(y}^2-2)\Big]\text{dy}$
$=\int\limits^2_\text{y}=-1\Big[\text{y}-\text{y}^2+2\Big]\text{dy}$
$=\Big[-\frac{\text{y}^3}{3}+\frac{\text{y}^2}{2}+2\text{y}\Big]^2_\text{y = -1}$
$=\frac{9}{2}$
First , we have to find the point of Intersection of the $2$ curves $y = y^2 - 2$
$y^2-y-2=0$
$y=-1, y=2$
Now , Area of the region bounded by the curve is
$=\int\limits^2_\text{y}=-1\Big[\text{y}-\text{(y}^2-2)\Big]\text{dy}$
$=\int\limits^2_\text{y}=-1\Big[\text{y}-\text{y}^2+2\Big]\text{dy}$
$=\Big[-\frac{\text{y}^3}{3}+\frac{\text{y}^2}{2}+2\text{y}\Big]^2_\text{y = -1}$
$=\frac{9}{2}$
MCQ 31 Mark
The area of the ellipse $\frac{\text{x}2}{9}+\frac{\text{y}^2}{4}=1$ in first quadrant is $6\pi$ 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 $\pi$ 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 $\pi$ sq. units.
- A$2$
- ✓$4$
- C$6$
- D$8$
Answer
View full question & answer→Correct option: B.
$4$
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
View full question & answer→Correct option: A.
$4$ sq. units
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$\text{None of these}$
Answer
View full question & answer→Correct option: B.
$1\text{ sq.}\text{ units}$
(B) $1\text{ sq.}\text{ units}$
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
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 $\text{y}_2=1-\text{x}\big)$
$= \int\limits^1_0\Big[\big(\sqrt{1-\text{x}^{2}}\big)-(1-\text{x})\Big]\text{dx}$
$=\int\limits^1_0\Big(\sqrt{1-\text{x}^2}-1+\text{x}\Big)\text{dx}$
$=\Big[\frac{\text{x}}{2}\sqrt{1-\text{x}^2}+\frac{1}{2}\sin^{-1}(\text{x})-\text{x}+\frac{\text{x}^2}{2}\Big]^1_0$
$=\bigg[\frac{1}{2}\sqrt{1-1^2}+\frac{1}{2}\sin^{-1}(1)-(1)+\frac{(1)^2}{2}\bigg]$
$-\bigg[\frac{(0)}{2}\sqrt{1-(0)^2}+\frac{1}{2}\sin^{-1}(0)-(0)+\frac{(0)^2}{2}\bigg]$
$=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)\text{square units}$
View full question & 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 $\text{y}_2=1-\text{x}\big)$
$= \int\limits^1_0\Big[\big(\sqrt{1-\text{x}^{2}}\big)-(1-\text{x})\Big]\text{dx}$
$=\int\limits^1_0\Big(\sqrt{1-\text{x}^2}-1+\text{x}\Big)\text{dx}$
$=\Big[\frac{\text{x}}{2}\sqrt{1-\text{x}^2}+\frac{1}{2}\sin^{-1}(\text{x})-\text{x}+\frac{\text{x}^2}{2}\Big]^1_0$
$=\bigg[\frac{1}{2}\sqrt{1-1^2}+\frac{1}{2}\sin^{-1}(1)-(1)+\frac{(1)^2}{2}\bigg]$
$-\bigg[\frac{(0)}{2}\sqrt{1-(0)^2}+\frac{1}{2}\sin^{-1}(0)-(0)+\frac{(0)^2}{2}\bigg]$
$=\Big(\frac{\pi}{4}-\frac{1}{2}\Big)\text{square units}$
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
View full question & 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
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 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 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 OACB given by (iv) - Area of triangle OAB by (v) $=(\pi-2)\text{ sq. units}$ Therefore, option (B) is correct.
$\Rightarrow\text{y}=\sqrt{2^2-\text{x}^2}\dots(\text{ii})$ Also, equation of the line is x + y = 2 ...(iii) Table of values
| $x$ | $0$ | $2$ |
| $y$ | $2$ | $0$ |
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
View full question & answer→Correct option: A.
$8\pi\text{ sq.}\text{units}$
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
View full question & answer→Correct option: B.
$2\sqrt{3}\text{ sq.}\text{ units}$
MCQ 101 Mark
The area bounded by the curve $2 x^2+y^2=2$ 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
View full question & answer→Correct option: B.
$\sqrt{2}\pi\text{ sq}.\text{units}$
MCQ 111 Mark
If $\text{y}=2\sin\text{x}+\sin2\text{x}$ for $0\leq \text{x}\leq 2\pi,$ then the area enclosed by the curve and $x-$ axis is :
- A$\frac{9}{2}\text{ sq.}\text{units}$
- B$8 \text{ sq. units}$
- ✓$12 \text{ sq. units}$
- D$4 \text{ sq. units}$
Answer
View full question & answer→Correct option: C.
$12 \text{ sq. units}$
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
$\text{Required area}=\int\limits^1_{-2}\text{y dx}$ $=\int\limits^1_{-2}\text{x}^3\text{dx}$ $=\Big[\frac{\text{x}^4}{4}\Big]^1_{-2}$ $=\Big[\frac14-\frac{(-2)^4}{4}\Big]$ $=\Big(\frac14-4\Big)=-\frac{15}{4}\text{ units}$ Thus, the correct answer is B.
View full question & 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.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
View full question & answer→Correct option: D.
None of these
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
The point of intersection of the curve and the straight line is $A(e, 1).$
Therefore, the area of the required region $\text{ABC},$
$\text{A} = \int\limits^1_0(\text{x}_1-\text{x}_2)\text{dy}$
$($where $, \text{x}_1 = \text{e}$ and $\text{x}_2 = \text{e}_{\text{y}})$
$= \int\limits^1_0(\text{e}-\text{e}^{\text{y}})\text{dy}$
$=\big [\text{ey}-\text{e}^{\text{y}}\big]^1_0$
$=\big\{\text{e}(1)-\text{e}^{(1)}\big\} -\big \{\text{e}(0)-\text{e}^{(0)}\big\}$
$= \text{e}-\text{e}+1$
$= 1 \text{ square unit}$
View full question & 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 $\text{ABC},$
$\text{A} = \int\limits^1_0(\text{x}_1-\text{x}_2)\text{dy}$
$($where $, \text{x}_1 = \text{e}$ and $\text{x}_2 = \text{e}_{\text{y}})$
$= \int\limits^1_0(\text{e}-\text{e}^{\text{y}})\text{dy}$
$=\big [\text{ey}-\text{e}^{\text{y}}\big]^1_0$
$=\big\{\text{e}(1)-\text{e}^{(1)}\big\} -\big \{\text{e}(0)-\text{e}^{(0)}\big\}$
$= \text{e}-\text{e}+1$
$= 1 \text{ square unit}$
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
View full question & answer→Correct option: C.
$3$ sq. units
MCQ 161 Mark
The area included between the parabolas $y^2=4 x$ and $x^2=4 y$ 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
View full question & answer→Correct option: C.
$\frac{16}{3}\text{ sq}\text{ unit}$
We know that, the area of region bounded by the parabolas $y^2=4 x$ and $x^2=4 y$ is :
$=\frac{16}{3}\text{ab}\text{ sq.}\text{ unit.}$
Therefore, $y^2=4 x$ and $x^2=4 y$ is :
$=\frac{16}{3}\text{ sq.}\text{ unit.}$
$(\because\text{a}=1,\text{b}=1)$
$=\frac{16}{3}\text{ab}\text{ sq.}\text{ unit.}$
Therefore, $y^2=4 x$ and $x^2=4 y$ is :
$=\frac{16}{3}\text{ sq.}\text{ unit.}$
$(\because\text{a}=1,\text{b}=1)$
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
View full question & answer→Correct option: A.
$20\pi\text{ sq}.\text{units}$
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
View full question & 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 $P\left(x, y_1\right)$ lies on $OA \ , O\left(x, y_2\right)$ lies on curve $O B$
Then, $y_1>0$
$ \Rightarrow\left|y_1\right|=y_1$
$\ y_2<0 $
$\Rightarrow\left|y_2\right|=-y_2$
Area of curve bound by the two lines $=$ shaded are $\text{(OADO)}\ +$ shaded area $\text{(OCBO)}$
$= \int\limits^0_{-2}|\text{y}_2|\text{dx}+\int\limits^1_0|\text{y}_1|\text{dx}$
$= \int\limits^0_{-2}-\text{y}_2\text{dx}+\int\limits^1_0\text{y}_1\text{dx}$
$=\int\limits^0_{-2}-(\text{x}^3)\text{dx}+\int\limits^1_0\text{x}^3\text{dx}$
$= \Big[-\frac{\text{x}^4}{4}\Big]^0_{-2}+\Big[\frac{\text{x}^4}{4}\Big]^1_0$
$= 0-\Big(-\frac{16}{4}\Big)+\Big(\frac{1}{4}-0\Big)$
$= 4+\frac{1}{4}$
$=\frac{17}{4}\text{ sq. units}$
If $P\left(x, y_1\right)$ lies on $OA \ , O\left(x, y_2\right)$ lies on curve $O B$
Then, $y_1>0$
$ \Rightarrow\left|y_1\right|=y_1$
$\ y_2<0 $
$\Rightarrow\left|y_2\right|=-y_2$
Area of curve bound by the two lines $=$ shaded are $\text{(OADO)}\ +$ shaded area $\text{(OCBO)}$
$= \int\limits^0_{-2}|\text{y}_2|\text{dx}+\int\limits^1_0|\text{y}_1|\text{dx}$
$= \int\limits^0_{-2}-\text{y}_2\text{dx}+\int\limits^1_0\text{y}_1\text{dx}$
$=\int\limits^0_{-2}-(\text{x}^3)\text{dx}+\int\limits^1_0\text{x}^3\text{dx}$
$= \Big[-\frac{\text{x}^4}{4}\Big]^0_{-2}+\Big[\frac{\text{x}^4}{4}\Big]^1_0$
$= 0-\Big(-\frac{16}{4}\Big)+\Big(\frac{1}{4}-0\Big)$
$= 4+\frac{1}{4}$
$=\frac{17}{4}\text{ sq. units}$
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
View full question & 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, $\int\limits\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n}}{\text{n+1}}+\text{c}$
$\Rightarrow\int\limits^5_2\text{x}^3\text{dx}=\Big[\frac{\text{x}^4}{4}\Big]^5_4$
$\Rightarrow\int\limits^5_2\text{x}^3\text{dx}=\frac{1}{4}(625-16)$
$=152.25$
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, $\int\limits\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n}}{\text{n+1}}+\text{c}$
$\Rightarrow\int\limits^5_2\text{x}^3\text{dx}=\Big[\frac{\text{x}^4}{4}\Big]^5_4$
$\Rightarrow\int\limits^5_2\text{x}^3\text{dx}=\frac{1}{4}(625-16)$
$=152.25$
MCQ 201 Mark
Area bounded by parabola $y^2=x$ and straight line $2y = x$ is :
- ✓$43$
- B$1$
- C$23$
- D$13$
Answer
View full question & 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 $2y = x$
$\Rightarrow y^2=2 y$
$\Rightarrow y^2-2 y=0$
$\Rightarrow y=0$ 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
$\text{A}=\int\limits_0^4(\text{y}_1-\text{y}_2)\text{ dx}$
$\Big[$Where, ${y}_1 =\sqrt{\text{x}}$ and ${y}_2=\frac{\text{x}}{2}\Big]$
$=\int\limits_0^4\Big(\sqrt{\text{x}}-\frac{\text{x}}{2}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}-\frac{1}{2}\times\frac{\text{x}^2}{2}\Bigg]_0^4$
$=\Big[\frac{2}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^2}{4}\Big]_0^4$
$=\frac{2}{3}4^\frac{3}{2}-\frac{1}{4}\times4^2-0$
$=\frac{2}{3}\times2^3-\frac{16}{4}$
$=\frac{16}{3}-4$
$=\frac{16-12}{3}$
$=\frac{4}{3}\text{ Sq units}$
$y^2=x$ and $2y = x$
$\Rightarrow y^2=2 y$
$\Rightarrow y^2-2 y=0$
$\Rightarrow y=0$ 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
$\text{A}=\int\limits_0^4(\text{y}_1-\text{y}_2)\text{ dx}$
$\Big[$Where, ${y}_1 =\sqrt{\text{x}}$ and ${y}_2=\frac{\text{x}}{2}\Big]$
$=\int\limits_0^4\Big(\sqrt{\text{x}}-\frac{\text{x}}{2}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}-\frac{1}{2}\times\frac{\text{x}^2}{2}\Bigg]_0^4$
$=\Big[\frac{2}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^2}{4}\Big]_0^4$
$=\frac{2}{3}4^\frac{3}{2}-\frac{1}{4}\times4^2-0$
$=\frac{2}{3}\times2^3-\frac{16}{4}$
$=\frac{16}{3}-4$
$=\frac{16-12}{3}$
$=\frac{4}{3}\text{ Sq units}$
MCQ 211 Mark
The area bounded by the curve $y^2=8 x$ and $x^2=8 y$ 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
Point of intersection of both the parabolas $y^2=8 x$ and $x^2=8 y$ 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$
$= \frac{2}{3} \times \sqrt{8} \times 8 \sqrt{8} - \frac{8^{2}}{3}$
$= \frac{2}{3} \times 8^{2} - \frac{8^{2}}{3}$
$=\frac{8^{2}}{3} (2 - 1)$
$=\frac{64}{3} \text{ sq units}$
View full question & answer→Correct option: A.
$\frac{16}{3}\text{ sq. units}$
Point of intersection of both the parabolas $y^2=8 x$ and $x^2=8 y$ 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$
$= \frac{2}{3} \times \sqrt{8} \times 8 \sqrt{8} - \frac{8^{2}}{3}$
$= \frac{2}{3} \times 8^{2} - \frac{8^{2}}{3}$
$=\frac{8^{2}}{3} (2 - 1)$
$=\frac{64}{3} \text{ sq units}$
MCQ 221 Mark
The area bounded by the parabola $y^2=4 a x$ and $x^2=4 a y$ 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
View full question & 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 $y^2=4 a x$
We get,
$\frac{\text{x}^4}{16\text{a}^{2}}=4\text{ax}$
$\Rightarrow x^4-64 a^3 x=0$
$\Rightarrow x\left(x^3-64 a^3\right)=0$
$\Rightarrow x=0$ or $x=4 a$
$\Rightarrow y=0$ or $y=4 a$
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 $\text{ y}_2=\frac{\text{x}^2}{\text{4a}}\Big)$
$= \int\limits^\text{4a}_0\Big(2\sqrt{\text{ax}}-\frac{\text{x}^2}{\text{4a}}\Big)\text{dx}$
$=\bigg[\frac{4\sqrt{\text{a}}}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^{3}}{12\text{a}}\bigg]^\text{4a}_0$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}(\text{4a}^\frac{3}{2})-\frac{(\text{4a})^3}{\text{12a}}\bigg]-\bigg[\frac{4\sqrt{\text{a}}}{3}(0)^\frac{3}{2}-\frac{(0)^3}{\text{12a}}\bigg]$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}\text{8a}^\frac{3}{2}-\frac{64\text{a}^3}{12\text{a}}\bigg]-0$
$= \frac{32\text{a}^2}{3}-\frac{16\text{a}^3}{3}$
$= \frac{16\text{a}^2}{3}\text{ square units}$
We get,
$\frac{\text{x}^4}{16\text{a}^{2}}=4\text{ax}$
$\Rightarrow x^4-64 a^3 x=0$
$\Rightarrow x\left(x^3-64 a^3\right)=0$
$\Rightarrow x=0$ or $x=4 a$
$\Rightarrow y=0$ or $y=4 a$
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 $\text{ y}_2=\frac{\text{x}^2}{\text{4a}}\Big)$
$= \int\limits^\text{4a}_0\Big(2\sqrt{\text{ax}}-\frac{\text{x}^2}{\text{4a}}\Big)\text{dx}$
$=\bigg[\frac{4\sqrt{\text{a}}}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^{3}}{12\text{a}}\bigg]^\text{4a}_0$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}(\text{4a}^\frac{3}{2})-\frac{(\text{4a})^3}{\text{12a}}\bigg]-\bigg[\frac{4\sqrt{\text{a}}}{3}(0)^\frac{3}{2}-\frac{(0)^3}{\text{12a}}\bigg]$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}\text{8a}^\frac{3}{2}-\frac{64\text{a}^3}{12\text{a}}\bigg]-0$
$= \frac{32\text{a}^2}{3}-\frac{16\text{a}^3}{3}$
$= \frac{16\text{a}^2}{3}\text{ square units}$
MCQ 231 Mark
The area bounded by the curve $x=3 y^2-9$ 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
View full question & answer→Correct option: A.
$8\text{sq.}\text{units}$
MCQ 241 Mark
Area lying between the curves $y^2=4 x$ and $y=2 x$ is :
- A$\frac{2}{3}$
- ✓$\frac{1}{3}$
- C$\frac{1}{4}$
- D$\frac{3}{4}$
Answer
View full question & 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=4 x $ and $y=2 x$
$\Rightarrow(2 x)^2=4 x$
$\Rightarrow 4 x^2=4 x$
$\Rightarrow x(x-1)=0$
$\Rightarrow x=0$ 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
$= \int\limits^2_0|\text{x}_2-\text{x}_1|\text{dy}$
$=\int\limits^2_0\Big[\Big(\frac{\text{y}}{2}\Big)-\Big(\frac{\text{y}^2}{4}\Big)\Big]\text{dy}$
$=\Big[\frac{1}{2}\Big(\frac{\text{y}^2}{2}\Big)-\frac{1}{4}\Big(\frac{\text{y}^3}{3}\Big)\Big]^2_0$
$=\frac{1}{4}(2)^2-\frac{1}{12}(2^3)-0$
$= 1 -\frac{8}{12}$
$= \frac{12-8}{12}$
$=\frac{1}{3}\text{ sq. units}$
$y^2=4 x $ and $y=2 x$
$\Rightarrow(2 x)^2=4 x$
$\Rightarrow 4 x^2=4 x$
$\Rightarrow x(x-1)=0$
$\Rightarrow x=0$ 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
$= \int\limits^2_0|\text{x}_2-\text{x}_1|\text{dy}$
$=\int\limits^2_0\Big[\Big(\frac{\text{y}}{2}\Big)-\Big(\frac{\text{y}^2}{4}\Big)\Big]\text{dy}$
$=\Big[\frac{1}{2}\Big(\frac{\text{y}^2}{2}\Big)-\frac{1}{4}\Big(\frac{\text{y}^3}{3}\Big)\Big]^2_0$
$=\frac{1}{4}(2)^2-\frac{1}{12}(2^3)-0$
$= 1 -\frac{8}{12}$
$= \frac{12-8}{12}$
$=\frac{1}{3}\text{ sq. units}$
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
View full question & answer→Correct option: B.
$\frac{1}{\text{a}} $
$=\text{Area} =\int\limits^\frac{\pi}{2}_0\cos\text{a x}\text{ dx}$
$=\Big[\frac{\sin\text{ax}}{\text{x}}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{0}-0=\frac{1}{\text{a}}$
$=\Big[\frac{\sin\text{ax}}{\text{x}}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{0}-0=\frac{1}{\text{a}}$
MCQ 261 Mark
The area bounded by the curve $y^2=8 x,$ 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
View full question & answer→Correct option: A.
$\frac{16}{3}$
$y^2=8 x$ 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
$\text{x} = 2\text{ area}\text{ OABO}= \int\limits^2_0|\text{y}|\text{dx}$
$= \int\limits^2_0\text{y}\text{ dx} $
$\{\text{y}>0, \Rightarrow|\text{y}|=\text{y}\}$
$= \int\limits^2_0\sqrt{8\text{x}}\text{ dx}$
$= 2\sqrt{2}\int\limits^2_0\sqrt{\text{xdx}}$
$= 2\sqrt{2}\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}\Bigg]^2_0$
$=2 \sqrt{2}\times \frac{2}{3}\Big(2^\frac{3}{2}-0\Big)$
$= 4 \frac{\sqrt{2}}{3}\times2\sqrt{2}$
$=\frac{16}{3}\text{ sq. units}$
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
$\text{x} = 2\text{ area}\text{ OABO}= \int\limits^2_0|\text{y}|\text{dx}$
$= \int\limits^2_0\text{y}\text{ dx} $
$\{\text{y}>0, \Rightarrow|\text{y}|=\text{y}\}$
$= \int\limits^2_0\sqrt{8\text{x}}\text{ dx}$
$= 2\sqrt{2}\int\limits^2_0\sqrt{\text{xdx}}$
$= 2\sqrt{2}\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}\Bigg]^2_0$
$=2 \sqrt{2}\times \frac{2}{3}\Big(2^\frac{3}{2}-0\Big)$
$= 4 \frac{\sqrt{2}}{3}\times2\sqrt{2}$
$=\frac{16}{3}\text{ sq. units}$
MCQ 271 Mark
Area of the region bounded by the curve $y^2=4 x, y-$ axis and the line $y = 3$, is :
- A$2$
- ✓$\frac{9}{4}$
- C$\frac{9}{3}$
- D$\frac{9}{2}$
Answer
$y^2=4 x$ represents a parabola with vertex at origin $O(0, 0)$ and symmetric about $+\text{ve x}-$ 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
$y^2=4 x$
$\Rightarrow 3^2=4 x$
$\Rightarrow\text{x}=\frac{9}{4}$
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,
$\text{ Area (OABO)}= \int\limits^3_0|\text{x}|\text{ dy}$
$= \int\limits^3_0\frac{\text{y}^2}{4}\text{dy}$
$=\Big[\frac{1}{4}\Big(\frac{\text{y}^2}{3}\Big)\Big]^3_0$
$= \frac{3^3}{12}$
$=\frac{9}{4}\text{ sq. units}$
View full question & answer→Correct option: B.
$\frac{9}{4}$
$y^2=4 x$ represents a parabola with vertex at origin $O(0, 0)$ and symmetric about $+\text{ve x}-$ 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
$y^2=4 x$
$\Rightarrow 3^2=4 x$
$\Rightarrow\text{x}=\frac{9}{4}$
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,
$\text{ Area (OABO)}= \int\limits^3_0|\text{x}|\text{ dy}$
$= \int\limits^3_0\frac{\text{y}^2}{4}\text{dy}$
$=\Big[\frac{1}{4}\Big(\frac{\text{y}^2}{3}\Big)\Big]^3_0$
$= \frac{3^3}{12}$
$=\frac{9}{4}\text{ sq. units}$
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
View full question & answer→Correct option: B.
$1$
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$2$ sq. units
- B$4$ sq. units
- C$3$ sq. units
- ✓$1$ sq. unit
Answer
View full question & answer→Correct option: D.
$1$ sq. unit
MCQ 301 Mark
The area bounded by the curve $x^2=4 y$ and straight line $x = 4y - 2$ is :
- A$\frac{3}{8}$
- B$\frac{5}{8}$
- C$\frac{7}{8}$
- ✓$\frac{9}{8}$
Answer
View full question & answer→Correct option: D.
$\frac{9}{8}$
The area bounded by the curve, $x^2=4 y$, 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 $\text{AL}$ and $\text{BM}$ 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, $=\Big(\frac{5}{6}+\frac{7}{24}\Big)=\frac{9}{8}\text{sq}.\text{ units}$
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 $\text{AL}$ and $\text{BM}$ 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, $=\Big(\frac{5}{6}+\frac{7}{24}\Big)=\frac{9}{8}\text{sq}.\text{ units}$
MCQ 311 Mark
The area bounded by the curve $y=x^4-2 x^3+x^2+3$ with $x-$ axis and ordinates corresponding to the minima of $y$ is :
- A$1$
- B$\frac{91}{30}$
- C$\frac{30}{9}$
- ✓$4$
Answer
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, $y=x^4-2 x^3+x^2+3)$
$= \int\limits^1_0(\text{x}^4-2\text{x}^3+\text{x}^2+3)\text{dx}$
$=\bigg[\frac{\text{x}^5}{5}-\frac{2\text{(x})^4}{4}+\frac{\text{x}^3}{3}+3\text{x}\bigg]^1_0$
$=\bigg[\frac{(1)^5}{5}-\frac{2(1)^4}{4}+\frac{(1)^3}{3}+3(1)\bigg]-\bigg[\frac{(0)^5}{5}-\frac{2(0)^4}{4}+\frac{(0)^3}{3}+3(0)\bigg]$
$=\big[\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3\big]- 0$
$=\frac{6-15+10+90}{3}$
$=\frac{91}{30}\text{ square units}$
View full question & 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, $y=x^4-2 x^3+x^2+3)$
$= \int\limits^1_0(\text{x}^4-2\text{x}^3+\text{x}^2+3)\text{dx}$
$=\bigg[\frac{\text{x}^5}{5}-\frac{2\text{(x})^4}{4}+\frac{\text{x}^3}{3}+3\text{x}\bigg]^1_0$
$=\bigg[\frac{(1)^5}{5}-\frac{2(1)^4}{4}+\frac{(1)^3}{3}+3(1)\bigg]-\bigg[\frac{(0)^5}{5}-\frac{2(0)^4}{4}+\frac{(0)^3}{3}+3(0)\bigg]$
$=\big[\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3\big]- 0$
$=\frac{6-15+10+90}{3}$
$=\frac{91}{30}\text{ square units}$
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
View full question & answer→Correct option: D.
None of these
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
View full question & answer→Correct option: D.
None of these
The line $\text{x} = \pi3$ meets the curve $\text{y} = \cos\text {x}\text{ at}\text { B}\pi\ 3,12$
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,
$= \text{A}_2 = \int\limits^\frac{\pi}{4}_0\text{y}_2\text{ dx}-\int\limits^\frac{\pi}{3}_\frac{\pi}{4}\text{y}_2\text{dx}$
$\big[$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
$\text{A}_1:\text{A}_2=\frac{\text{A}_1}{\text{A}_2}=\frac{\frac{\sqrt{3}}{2}}{\frac{4-\sqrt{3}}{4}}=\frac{2\sqrt{3}}{4-\sqrt{3}}$
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,
$= \text{A}_2 = \int\limits^\frac{\pi}{4}_0\text{y}_2\text{ dx}-\int\limits^\frac{\pi}{3}_\frac{\pi}{4}\text{y}_2\text{dx}$
$\big[$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
$\text{A}_1:\text{A}_2=\frac{\text{A}_1}{\text{A}_2}=\frac{\frac{\sqrt{3}}{2}}{\frac{4-\sqrt{3}}{4}}=\frac{2\sqrt{3}}{4-\sqrt{3}}$
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
View full question & answer→Correct option: C.
$\frac{1}{6}$
Required area $=\text{A}=\int\limits^1_0\big(\sqrt{\text{x}}-\text{x}\big)\text{dx}$
$=\Big[\frac{2}{3}\text{x}^{\frac{3}{2}}-\frac{\text{x}^2}{2}\Big]^1_0$
$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$
$=\Big[\frac{2}{3}\text{x}^{\frac{3}{2}}-\frac{\text{x}^2}{2}\Big]^1_0$
$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$
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
View full question & answer→Correct option: A.
$2\text{ sq. units}$
$\text{A}=\int^\limits{\pi}_0\text{y}\text{ dx}$
$=\int^\limits{\pi}_0\sin(\text{x})\text{dx}$
$=\big[-\cos(\text{x})\big]^{\pi}_0$
$= -\cos(\pi)+\cos(0)$
$= 1 + 1$
$= 2 \text{ square units}$
$=\int^\limits{\pi}_0\sin(\text{x})\text{dx}$
$=\big[-\cos(\text{x})\big]^{\pi}_0$
$= -\cos(\pi)+\cos(0)$
$= 1 + 1$
$= 2 \text{ square units}$
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
View full question & 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
$\text{A}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x dx}$
$=-\Big[\cos\text{x}\Big]^{\frac{\pi}{2}}_0=-\Big[\cos\frac{\pi}{2}-\cos0\Big]$
$=-[0-1]=1\text{ sq. units}$
$\text{A}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x dx}$
$=-\Big[\cos\text{x}\Big]^{\frac{\pi}{2}}_0=-\Big[\cos\frac{\pi}{2}-\cos0\Big]$
$=-[0-1]=1\text{ sq. units}$
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
View full question & answer→Correct option: B.
$4$
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$\sin(3\text{x}+4)$
- ✓$\sin(3\text{x}+4)+3(\text{x}-1)\cos(3\text{x}+4)$
- Dnone of these
Answer
View full question & 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$)$
$\Rightarrow \text{f}(\text{x})=\frac{\text{d}}{\text{dx}}((\text{x} - 1)\sin(3\text{x}+ 4))$
$= \sin (3\text{x} + 4)\frac{\text{d}}{\text{dx}}(\text{x} - 1) + (\text{x} - 1)\frac{\text{d}}{\text{dx}}\sin(3\text{x} + 4)$
$= \sin (3\text{x} + 4) + 3(\text{x} - 1)\cos(3\text{x} + 4)$
$y = fx$
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$)$
$\Rightarrow \text{f}(\text{x})=\frac{\text{d}}{\text{dx}}((\text{x} - 1)\sin(3\text{x}+ 4))$
$= \sin (3\text{x} + 4)\frac{\text{d}}{\text{dx}}(\text{x} - 1) + (\text{x} - 1)\frac{\text{d}}{\text{dx}}\sin(3\text{x} + 4)$
$= \sin (3\text{x} + 4) + 3(\text{x} - 1)\cos(3\text{x} + 4)$
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
View full question & answer→Correct option: D.
$4\ sq$ units
$(d),$ as $\text{x}=\sin$ is positive in $1^{st}$ and $2^{nd}$ quadrant and negative is $3^{rd}$ and $4^{th}$ quadrant.
$=$Area$=\int\limits^{2\pi}_0\sin\text{x}\text{ dx}$
$=\int\limits^\pi_0\sin\text{x}+\int\limits^{2\pi}_\pi(-\sin\text{x})\text{dx}$
$=4\text{ sq}$ units
$=$Area$=\int\limits^{2\pi}_0\sin\text{x}\text{ dx}$
$=\int\limits^\pi_0\sin\text{x}+\int\limits^{2\pi}_\pi(-\sin\text{x})\text{dx}$
$=4\text{ sq}$ units
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$\frac{\pi}{6}+\frac{\sqrt{3}+1}{8}$
- ✓$\frac{\pi}{6}-\frac{\sqrt{3}-1}{8}$
- Dnone of these
Answer
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.
$x^2+y^2-6 x-4 y+12=0......(i)$
$y=x \ldots(ii)$
$\text{x} = \frac{5}{2}\ ...(\text{iii})$
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,
$x^2+y^2-6 x-4 y+12=0$
$\Rightarrow(x-3)^2+(y-2)^2=1$
$\Rightarrow(y-2)^2=1-(x-3)^2$
$\Rightarrow\text{y}-2 = \pm\sqrt{1-(\text{x}-3)^{2}}$
$\Rightarrow\text{y} = \pm\sqrt{1-(\text{x}-3})^{2}+2$
$\Rightarrow\text{y} = \sqrt{1-(\text{x}-3)^{2}}+2$ 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 $\text{ABC},$
$\text{A}= \int\limits^\frac{5}{2}_{2}(\text{y}_2-\text{y}_1)\text{ dx} \big($where$,\text{ y}_1 = -\sqrt{1-(\text{x}-3)^2} +2$ and $\text{ y}^2 = \text{x}\big)$
$= \int\limits^\frac{5}{2}_{2}\big[\text {x}-\big(-\sqrt{1(\text{x}-3)^{2}}+2\big)\big]\text{dx}$
$= \int^\limits\frac{5}{2}_2\big[\text{x}+\sqrt{1-(\text{x}-3)^{2}}\big]\text{dx}$
$= \bigg[\frac{\text{x}^{2}}{2}+\frac{(\text{x}-3)}{2}\sqrt{1-(\text{x}-3)^{2}}+\frac{1}{2}\sin^{-1}(\text{x}-3)-2\text{x}\bigg]^{\frac{5}{2}}_{2}$
$=\Bigg[\frac{\big(\frac{5}{2}\big)^{2}}{2}+\frac{\frac{5}{2}-3}{2}\sqrt{1-\Big\{\Big(\frac{5}{2}\Big)-3\Big\}^2}+\frac{1}{2}\sin^{-1}=\Big(\frac{5}{2}-3\Big)-2\Big(\frac{5}{2}\Big)\Bigg]$
$-\bigg[\frac{2^{2}}{2}+\frac{2-3}{2}\sqrt{1-(2-3)^2}+\frac{1}{2}\sin^{-1}(2-3)-2(2)\bigg]$
$= \bigg[\frac{25}{8}-\frac{1}{4}\sqrt{1-\frac{1}{4}}+\frac{1}{2}\sin^{-1}\Big(-\frac{1}{2}\Big)-5\bigg]$
$- \bigg[2-\frac{1}{2}\times0+\frac{1}{2}\sin^{-1}(-1)-4\bigg]$
$= \Big[-\frac{15}{8}-\frac{\sqrt{3}}{8}+\frac{1}{2}\times\Big(-\frac{\pi}{6}\Big)\Big]-\Big[+\frac{1}{2}\times\Big(-\frac{\pi}{2}\Big)-2\Big]$
$= -\frac{15}{8}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}+2$
$=\frac{\pi}{6}-\frac{\sqrt{3}-1}{8}$
View full question & 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.
$x^2+y^2-6 x-4 y+12=0......(i)$
$y=x \ldots(ii)$
$\text{x} = \frac{5}{2}\ ...(\text{iii})$
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,
$x^2+y^2-6 x-4 y+12=0$
$\Rightarrow(x-3)^2+(y-2)^2=1$
$\Rightarrow(y-2)^2=1-(x-3)^2$
$\Rightarrow\text{y}-2 = \pm\sqrt{1-(\text{x}-3)^{2}}$
$\Rightarrow\text{y} = \pm\sqrt{1-(\text{x}-3})^{2}+2$
$\Rightarrow\text{y} = \sqrt{1-(\text{x}-3)^{2}}+2$ 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 $\text{ABC},$
$\text{A}= \int\limits^\frac{5}{2}_{2}(\text{y}_2-\text{y}_1)\text{ dx} \big($where$,\text{ y}_1 = -\sqrt{1-(\text{x}-3)^2} +2$ and $\text{ y}^2 = \text{x}\big)$
$= \int\limits^\frac{5}{2}_{2}\big[\text {x}-\big(-\sqrt{1(\text{x}-3)^{2}}+2\big)\big]\text{dx}$
$= \int^\limits\frac{5}{2}_2\big[\text{x}+\sqrt{1-(\text{x}-3)^{2}}\big]\text{dx}$
$= \bigg[\frac{\text{x}^{2}}{2}+\frac{(\text{x}-3)}{2}\sqrt{1-(\text{x}-3)^{2}}+\frac{1}{2}\sin^{-1}(\text{x}-3)-2\text{x}\bigg]^{\frac{5}{2}}_{2}$
$=\Bigg[\frac{\big(\frac{5}{2}\big)^{2}}{2}+\frac{\frac{5}{2}-3}{2}\sqrt{1-\Big\{\Big(\frac{5}{2}\Big)-3\Big\}^2}+\frac{1}{2}\sin^{-1}=\Big(\frac{5}{2}-3\Big)-2\Big(\frac{5}{2}\Big)\Bigg]$
$-\bigg[\frac{2^{2}}{2}+\frac{2-3}{2}\sqrt{1-(2-3)^2}+\frac{1}{2}\sin^{-1}(2-3)-2(2)\bigg]$
$= \bigg[\frac{25}{8}-\frac{1}{4}\sqrt{1-\frac{1}{4}}+\frac{1}{2}\sin^{-1}\Big(-\frac{1}{2}\Big)-5\bigg]$
$- \bigg[2-\frac{1}{2}\times0+\frac{1}{2}\sin^{-1}(-1)-4\bigg]$
$= \Big[-\frac{15}{8}-\frac{\sqrt{3}}{8}+\frac{1}{2}\times\Big(-\frac{\pi}{6}\Big)\Big]-\Big[+\frac{1}{2}\times\Big(-\frac{\pi}{2}\Big)-2\Big]$
$= -\frac{15}{8}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}+2$
$=\frac{\pi}{6}-\frac{\sqrt{3}-1}{8}$
MCQ 411 Mark
Choose the correct answer from the given four options: The area of the region bounded by the curve $x^2=4 y$ and the straight line $x = 4y - 2$ is:
- A$\frac{3}{8} sq.$ units
- B$\frac{5}{8}\text{ sq.}$ units
- C$\frac{7}{8}\text{ sq.}$ units
- ✓$\frac{9}{8}\text{ sq.}$ units
Answer
View full question & answer→Correct option: D.
$\frac{9}{8}\text{ sq.}$ units
We have parabola $x^2=4 y$ 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 $x^2=4 y$ and $x=4 y-2$ 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
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 $x^2=4 y$ and $x=4 y-2$ 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
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
View full question & answer→Correct option: A.
$1\ sq.$ unit
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
View full question & 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
$\Big(\frac{\text{x}}{2}\Big)^2=\text{x}$
$\Rightarrow\ \text{x}^2=4^{\frac{3}{2}}$
$\Rightarrow\ \text{x(x}-4)=0$
$\Rightarrow\ \text{x}=4,0$
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
MCQ 441 Mark
The area bounded by the lines $y = |x| - 1$ and $y = -|x| + 1$ is:
- A$1\ sq.$ unit
- ✓$2\ sq.$ unit
- C$2\sqrt{2}\text{ sq}.$ units
- D$4\ sq.$ units
Answer
View full question & answer→Correct option: B.
$2\ sq.$ unit
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
View full question & answer→Correct option: A.
$2\ sq.$ units
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
View full question & 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
$=\sqrt{2}-(\sqrt{\text{x}})^2+3=0$
$\Rightarrow(\sqrt{\text{x}})^2-2\sqrt{\text{x}}-3=0$
$\Rightarrow(\sqrt{\text{x}}-3)(\sqrt{\text{x}}-3=0$
$\Rightarrow\sqrt{\text{x}}-3$
$\because\sqrt{\text{x}}=-1$ is not possible
$\therefore\text{y}=3$
Hence required area
$=\int\limits^3_0(\text{x}_2-\text{x}_1\text{dy})$
$=\int\limits^3_0(2\text{y}+3)-\text{y}^2)\text{dy}$
$=\Big[\text{y}^2+3\text{y}-\frac{\text{y}^3}{3}\Big]^3_0$
$=9+9-9=9$
Solving $(1)$ and $(2),$ we get
$=\sqrt{2}-(\sqrt{\text{x}})^2+3=0$
$\Rightarrow(\sqrt{\text{x}})^2-2\sqrt{\text{x}}-3=0$
$\Rightarrow(\sqrt{\text{x}}-3)(\sqrt{\text{x}}-3=0$
$\Rightarrow\sqrt{\text{x}}-3$
$\because\sqrt{\text{x}}=-1$ is not possible
$\therefore\text{y}=3$
Hence required area
$=\int\limits^3_0(\text{x}_2-\text{x}_1\text{dy})$
$=\int\limits^3_0(2\text{y}+3)-\text{y}^2)\text{dy}$
$=\Big[\text{y}^2+3\text{y}-\frac{\text{y}^3}{3}\Big]^3_0$
$=9+9-9=9$
MCQ 471 Mark
The area bounded by the line $y = 2x – 2, y = – x$ and $x-$axis is given by:
- A$\frac{9}{2}\text{ sq.}$ units
- B$\frac{43}{6}\text{ sq.}$ units
- C$\frac{35}{6}\text{ sq.}$ units
- ✓None of these
Answer
View full question & answer→Correct option: D.
None of these
MCQ 481 Mark
The area bounded by $y –1 = |x|, y = 0$ and $|x| =\frac{1}{2}$ will be:
- A$\frac{3}{4}$
- B$\frac{3}{2}$
- ✓$\frac{5}{4}$
- DNone of these
Answer
View full question & answer→Correct option: C.
$\frac{5}{4}$
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
View full question & 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
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
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
We have$, Y-$axis i.e.$, x = 0, \text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x},$
where $0\leq\text{x}\leq\frac{\pi}{2}$
View full question & 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 $0\leq\text{x}\leq\frac{\pi}{2}$