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
$\frac{\pi\text{ab}}{2}\text{ sq}.\text{units}$
AnswerCorrect option: C. $\pi\text{ ab}\text{ sq}.\text{units} $
View full question & answer→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
$\frac{9\sqrt{17}}{6}\text{sq}.\text{units}$
- ✓
$\frac{17\sqrt{17}}{6}\text{sq}.\text{units}$
AnswerCorrect option: D. $\frac{17\sqrt{17}}{6}\text{sq}.\text{units}$
View full question & answer→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
$\frac{\pi^3-4}{4\text{ sq.}\text{ units}}$
AnswerCorrect option: A. $\frac{\pi^3-8}{4\text{ sq.}\text{ units}}$
View full question & answer→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
$\frac{4}{3}$
AnswerCorrect option: C. $\frac{2}{3}$
The given equation of the curve is
$y = x|x|$
$\Rightarrow \text{y}= \begin{cases}\text{x}^2 & \text{x} \geq0\\\text{-x}^2 & \text{x} < 0\end{cases}$
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 $P\left(x, y_1\right), x>0$ is a point on $y=x|x|$ then $y_1 > 0 \Rightarrow\left|y_1\right|=y_1$
And $Q\left(x, y_2\right), x<0$ is a point on $y=x|x|$ then $y_2<0 \Rightarrow\left|y_2\right|=-y_2$
Required area $=\int\limits^0_{-1}|\text{y}_2|\text{dx}+\int\limits^1_0|\text{y}_1|\text{dx}$
$= \int\limits^0_{-1}-\text{y}_2\text{ dx}+\int\limits^1_0\text{y}_1\text{ dx}$
$= \int\limits^0_{-1}-(-\text{x}^{2})\text{dx}+\int\limits^1_0\text{x}^2\text{ dx}$
$=\int\limits^0_{-1}\text{x}^2\text{dx}+\int\limits^1_0\text{x}^2\text{dx}$
$= \Big[\frac{\text{x}^3}{3}\Big]^0_{-1} +\Big[\frac{\text{x}^3}{3}\Big]^1_0$
$= \bigg[0-\frac{(-1)^3}{3}\bigg]+\Big(\frac{1^3}{3}-0\Big)$
$= \frac{1}{3}+\frac{1}{3}$
$= \frac{2}{3}\text{ sq. units}$ View full question & answer→MCQ 1551 Mark
Area bounded by the curve $\text{y}=\log\text{x}$ and the coordinate axes is:
View full question & answer→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
$\frac{13}{2}\text{ sq. units}$
AnswerCorrect option: A. $\frac{7}{2}\text{ sq. units}$
From the figure, are of the shaded region,
$\text{A}=\int\limits^{3}_{2}(\text{x}+1)\text{dx}=\bigg[\frac{\text{x}^2}{2}+\text{x}\bigg]^3_2$
$=\bigg[\frac{9}{2}+3-\frac{4}{2}-2\bigg]$
$=\frac{7}{2}\text{sq. units}$ View full question & answer→MCQ 1571 Mark
The area bounded by the curve $y=x^2+4 x+5$, 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
AnswerCorrect option: B. $4\frac{2}{3}\ \text{sq}.\text{ units}$
View full question & answer→MCQ 1581 Mark
The area of the region bounded by the curve $\text{y}=\sqrt{16-\text{x}^2}$ and $\text{ x}-$ axis is :
- ✓
$8\text{p sq. units}$
- B
$20\text{p sq. units}$
- C
$16\text{p sq. units}$
- D
$256\text{p sq. units}$
AnswerCorrect option: A. $8\text{p sq. units}$
View full question & answer→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
$\frac{2}{3}(\pi-2)$
AnswerCorrect option: A. $3(\pi-2)$
View full question & answer→MCQ 1601 Mark
The area of the region bounded by the curve $x = 2y + 3$ and the lines $y = 1$ and $y = -1$ is :
AnswerCorrect option: C. $6 \text{ sq. units}$
View full question & answer→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
$\sqrt{2}$
AnswerCorrect option: B. $\sqrt{2}-1$
Points of intersection is obtained by solving $y = \sin x$ 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 $0\leq\text{x}\leq\pi2$
$\text{A} = \int\limits^\frac{1}{\sqrt{2}}_0|\text{x}_1|\text{dy}+\int\limits^1_\frac{1}{\sqrt{2}}|\text{x}_2|\text{dy}$
$=\int\limits^\frac{1}{\sqrt{2}}_0\text{x}_1\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\text{x}_2\text{ dy}$
$= \int\limits^\frac{1}{\sqrt{2}}_0\sin^{-1}\text{y}\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\cos^{-1}\text{y}\text{ dy}$
$= \Big[\text{y}\sin^{-1}\text{y}+\sqrt{1-\text{y}^2}\Big]^\frac{1}{\sqrt{2}}_0+\Big[\text{y}\cos^{-1}\text{y}-\sqrt{1-\text{y}^2}\Big]^1_\frac{1}{\sqrt{2}}$
$= \Big[\frac{1}{\sqrt{2}}\sin^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}-1\Big]$
$+\bigg[1\times\cos^{-1}-0-\frac{1}{\sqrt{2}}\cos^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}\bigg]$
$= \frac{2}{\sqrt{2}}-1$
$=\big(\sqrt{2}-1\big)\text{sq. units}$
View full question & answer→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
$\sqrt2.$
AnswerCorrect 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. View full question & answer→MCQ 1631 Mark
The area bounded by the circles $= 4 x^2+y^2=1, x^2+y^2=4$ in the first Quadrant is :
- A
$\frac{\pi}{2}$
- ✓
$\frac{3\pi}{4}$
- C
$3\pi$
- D
$\frac{\pi}{4}$
AnswerCorrect 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 $=\frac{3\pi}{4}$
View full question & answer→MCQ 1641 Mark
Let $f(x) = x^2-3 x+2$ 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
AnswerCorrect 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 $f(∣x∣)$
$=\int\limits^1_{-1}\text{f(dx)}=2$
$=\int\limits^1_0\text{f(dx)}=2,\frac{5}{6}=\frac{5}{3}\text{sq}\text{ units}$
View full question & answer→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
$\frac{\pi}{2}+\frac{4}{3}$
AnswerCorrect option: C. $\frac{\pi}{2}-\frac{4}{3}$
The required area
$=2\int\limits^1_0\sqrt{1-\text{x}}\text{dy}+2.\Big(\frac{1}{4}.\pi.1^2\Big)$
$=-2\Bigg[\frac{2(1-\text{x)}^\frac{3}{2}}{3}\Bigg]^1_0+\frac{\pi}{2}$
$=\frac{\pi}{2}-\frac{4}{3}$
View full question & answer→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
$\frac{1}{4}$
AnswerCorrect 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 $=1\times1\times\frac{1}{2}$
View full question & answer→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
$\frac{5}{18}\ \text{sq.}\text{units}$
- ✓
$\frac{9}{2}\ \text{sq.}\text{units}$
AnswerCorrect option: D. $\frac{9}{2}\ \text{sq.}\text{units}$
View full question & answer→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 :
AnswerCorrect 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} $
$=\frac{16}{5}\bigg[0+\frac{5^2}{2}\sin^{-1}1-0-0\bigg]$
$=\frac{16}{5}.\frac{25}{2}.\frac{\pi}{2}=20\pi\text{ sq. units}$ View full question & answer→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
$8\text{ sq}\text{ units}$
AnswerCorrect option: C. $6\text{ sq}\text{ units}$
as area $=\int\limits^1_0(2\text{y}+3)\text{dy}$
$=6\text{ sq}\text{ units}$
View full question & answer→MCQ 1701 Mark
The area of the region bounded by the line $y = | x - 2 |, x = 1, x = 3$ and $x-$ axis is :
- A
$4 \text{ sq. units}$
- B
$2 \text{ sq. units}$
- C
$3 \text{ sq. units}$
- ✓
$1 \text{ sq. unit}$
AnswerCorrect option: D. $1 \text{ sq. unit}$
View full question & answer→MCQ 1711 Mark
The area of the region bounded by $y = (x - 4)^2, y = 16 - x^2 $ and the $x$ axis, is:
- A
$\frac{68}{4}$
- B
$\frac{64}{5}$
- ✓
$\frac{64}{3}$
- D
$\frac{64}{9}$
AnswerCorrect option: C. $\frac{64}{3}$
$\text{y}_1 = \text{x}^2 + 16 - 8\text{x}, \text{y}^2 = 16 - \text{x}^2 $
$16 - \text{x}^2 = \text{x}^2 = 16 - 8\text{x}$
$\Rightarrow 2\text{x}^2 - 8\text{x} = 0$
$\text{x}^2 - 4 =0, \text{x} = 0,4$
$=\int\limits^4_0(\text{y}_1-\text{y}_2)\text{ dx}$
$\Rightarrow\int\limits_0(2\text{x}^2-8\text{x})\text{ dx}$
$\Rightarrow\Big[\frac{2\text{x}^3}{3}-4\text{x}\Big]^4_0$
$\frac{128}{3}-64=\frac{64}{3}$
View full question & answer→MCQ 1721 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:
View full question & answer→MCQ 1731 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
$\frac{\pi\text{ab}}{2}\text{ sq.}\text{ units}$
AnswerCorrect option: C. $\pi\text{ ab}\text{ sq.}\text{ units}$
View full question & answer→MCQ 1741 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
AnswerCorrect option: A. $\frac{45}{4}$
View full question & answer→MCQ 1751 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
AnswerCorrect option: B. $8$ units
View full question & answer→MCQ 1761 Mark
The area bounded by the curve $y=\sqrt{x}, Y$-axis and between the lines $y=0$ and $y=3$ is
View full question & answer→MCQ 1771 Mark
Shown below is the curve defined by the equation $y=\log (x+1)$ for $x \geq 0$.
Which of these is the area of the shaded region? - ✓
$6 \log (2)-2$
- B
$6 \log (2)-6$
- C
$6 \log (2)$
- D
$5 \log (2)$
AnswerCorrect option: A. $6 \log (2)-2$
$6 \log (2)-2$
View full question & answer→MCQ 1781 Mark
The area of the region bounded by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is
- ✓
$20 \pi$ sq. units
- B
$20 \pi^2$ sq. units
- C
$16 \pi^2$ sq. units
- D
$25 \pi$ sq. units
AnswerCorrect option: A. $20 \pi$ sq. units
(a) : Area of the region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi a b$ sq. units
$\therefore \quad$ Required area $=\pi \times 5 \times 4=20 \pi$ sq. units
View full question & answer→MCQ 1791 Mark
The area of the region bounded by the curve $y=\sin x$ between the ordinates $x=0, x=\frac{\pi}{2}$ and the $x$-axis is
Answer(d): We have, $y=\sin x, 0 \leq x \leq \frac{\pi}{2}$
$\therefore \quad$ Required area
$
=\int_0^{\pi / 2} \sin x d x=[-\cos x]_0^{\pi / 2}=-[0-1]=1 \text { sq. unit }
$ View full question & answer→MCQ 1801 Mark
The area of the region bounded by parabola $y^2=x$ and the straight line $2 y=x$ is
- ✓
$\frac{4}{3}$ sq. units
- B
$1$ sq. unit
- C
$\frac{2}{3}$ sq. unit
- D
$\frac{1}{3}$ sq. unit
AnswerCorrect option: A. $\frac{4}{3}$ sq. units
We have $2 y=x \ldots (i),$ a straight line, and $y^2=x...(ii),$ a parabola with vertex $(0,0)$.
Solving $(i)$ and $(ii),$ we get $x=0$ and $x=4$.
$\therefore$ Required area
$=\int_0^4\left(\sqrt{x}-\frac{x}{2}\right) d x=\left[\frac{2}{3} x^{3 / 2}-\frac{x^2}{4}\right]_0^4=\frac{2}{3} \times 8-\frac{16}{4}$
$=\frac{16-12}{3}=\frac{4}{3} \text { sq. units }$ View full question & answer→MCQ 1811 Mark
Area of the region bounded by the curve $y=\cos x$ between $x=0$ and $x=\pi$ is
Answer(a): We have, $y=\cos x$
$\therefore \quad$ Required area
$
=2 \int_0^{\pi / 2} \cos x d x=2[\sin x]_0^{\pi / 2}=2 \text { sq. units }
$ View full question & answer→MCQ 1821 Mark
The area bounded by the $x$-axis, the curve $y=f(x)$ and the lines $x=1, x=b$ is equal to $\sqrt{b^2+1}-\sqrt{2}$ for all $b>1$. Which of the following can be $f(x)$ ?
- A
$\sqrt{x-1}$
- B
$\sqrt{x+1}$
- C
$\sqrt{x^2+1}$
- ✓
$x / \sqrt{x^2+1}$
AnswerCorrect option: D. $x / \sqrt{x^2+1}$
(d) : We have, $\int_1^b f(x) d x=\sqrt{b^2+1}-\sqrt{2}$ On differentiating w.r.t. $b$, we get
$
f(b)=\frac{2 b}{2 \sqrt{b^2+1}} \Rightarrow f(x)=\frac{x}{\sqrt{x^2+1}}
$
View full question & answer→MCQ 1831 Mark
The area bounded by the curve $x^2=4 y+4$ and line $3 x+4 y=0$ is
- A
$\frac{25}{4}$ sq. units
- B
$\frac{125}{8} sq$. units
- C
$\frac{125}{16}$ sq. units
- ✓
$\frac{125}{24}$ sq. units
AnswerCorrect option: D. $\frac{125}{24}$ sq. units
(d): We have, $x^2=4 y+4$ ...(i)
and $3 x+4 y=0$ ...(ii)
Solving (i) and (ii), we get $x=-4,1$
$\therefore \quad$ Required area
$
=\int_{-4}^1\left(-\frac{3 x}{4}-\frac{x^2}{4}+1\right) d x=\left[-\frac{3}{8} x^2-\frac{x^3}{12}+x\right]_{-4}^1
$
$=-\frac{3}{8}(1-16)-\frac{1}{12}(1+64)+5=\frac{45}{8}-\frac{5}{12}=\frac{125}{24}$ sq. units View full question & answer→MCQ 1841 Mark
The area bounded by the curve $y=\sec ^2 x, y=0$ and $|x|=\frac{\pi}{3}$ is
- A
$\sqrt{3}$ sq. units
- B
$\sqrt{2}$ sq. units
- ✓
$2 \sqrt{3}$ sq. units
- D
AnswerCorrect option: C. $2 \sqrt{3}$ sq. units
(c): We have, $y=\sec ^2 x$ and $y=0$ and $x=\frac{\pi}{3},-\frac{\pi}{3}$
$\therefore \quad$ Required area
$
=\int_{-\pi / 3}^{\pi / 3} \sec ^2 x d x=[\tan x]_{-\pi / 3}^{\pi / 3}=2 \sqrt{3} \text { sq. units }
$ View full question & answer→MCQ 1851 Mark
Area of the region bounded by the curve $y=\tan x$, line $x=\frac{\pi}{4}$ and the $x$-axis is
AnswerCorrect option: B. $\frac{1}{2} \log 2$ sq. units
We have, $y=\tan x$ and $x=\frac{\pi}{4}$
$\therefore$ Required area
$=\int_0^{\pi / 4} \tan x d x=[-\log |\cos x|]_0^{\pi / 4}=-\log \frac{1}{\sqrt{2}}+\log 1$
$=\log \sqrt{2}=\frac{1}{2} \log 2 \text { sq. units }$ View full question & answer→MCQ 1861 Mark
Area bounded by the curve $y=\cos x$ between $x=0$ and $x=\frac{3 \pi}{2}$ is
- A
$1$ sq. unit
- B
$2$ sq. units
- ✓
$3$ sq. units
- D
$4$ sq. units
AnswerCorrect option: C. $3$ sq. units
We have, $y=\cos x$, whose graph is shown below, between $x=0$ and $x=\frac{3 \pi}{2}$
$\therefore \quad$ Required area
$=\int_0^{\pi / 2} \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|$
$=[\sin x]_0^{\pi / 2}+\left|[\sin x]_{\pi / 2}^{3 \pi / 2}\right|$
$=1+|(-1-1)|$
$=1+2$
$=3 \text { sq. units }$ View full question & answer→MCQ 1871 Mark
The area bounded by the curve $x^2+y^2=1$ in first quadrant is
- ✓
$\frac{\pi}{4}$ sq. units
- B
$\frac{\pi}{2}$ sq. units
- C
$\frac{\pi}{3}$ sq. units
- D
$\frac{\pi}{6}$ sq. units
AnswerCorrect option: A. $\frac{\pi}{4}$ sq. units
We have, $x^2+y^2=1$, which is a circle with centre $(0,0)$ and radius $=1$.
Required area
$=\int_0^1 \sqrt{1-x^2} d x$
$=\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} \frac{x}{1}\right]_0^1$
$=\left[\frac{1}{2} \sin ^{-1} 1\right]$
$=\left(\frac{1}{2} \times \frac{\pi}{2}\right)$
$=\frac{\pi}{4} \text { sq. units }$ View full question & answer→MCQ 1881 Mark
Find the area enclosed by the parabola $y^2=x$ and the line $y+x=2$ and the $x-$axis.
- A
$\frac{5}{6}$ sq. units
- ✓
$\frac{7}{6}$ sq. units
- C
$\frac{6}{7}$ sq. units
- D
$\frac{4}{7}$ sq. units
AnswerCorrect option: B. $\frac{7}{6}$ sq. units
The given line and parabola meet at the points $(1,1)$ and $(4,-2)$.
$\therefore$ Required area
$=\int_0^1 \sqrt{x} d x+\int_1^2(2-x) d x$
$=\left[\frac{x^{3 / 2}}{3 / 2}\right]_0^1+\left[2 x-\frac{x^2}{2}\right]_1^2$
$=\frac{2}{3}(1-0)+\left(2 \times 2-\frac{2^2}{2}\right)-\left(2-\frac{1}{2}\right)$
$=\frac{2}{3}+2-\frac{3}{2}$
$=\frac{4+12-9}{6}$
$=\frac{7}{6}$ sq. units View full question & answer→MCQ 1891 Mark
Find the area above $x$-axis, bounded by the curves $y=2^{k x}, x=0$ and $x=2$.
- ✓
$\frac{4^k-1}{k \log _e 2}$
- B
$\frac{2^k-1}{2 \log _e 2}$
- C
$\frac{3-k}{k \log _e 2}$
- D
$\frac{-1+3^k}{2 \log _e 2}$
AnswerCorrect option: A. $\frac{4^k-1}{k \log _e 2}$
(a) : Required area $=\int_0^2 y d x$
$
=\int_0^2 2^{k x} d x=\left[\frac{2^{k x}}{k \log _e 2}\right]_0^2=\frac{2^{2 k}}{k \log _e 2}-\frac{1}{k \log _e 2}=\frac{4^k-1}{k \log _e 2}
$ View full question & answer→MCQ 1901 Mark
The area bounded by the curve $x=3 y^2-9$ and the line $x=0, y=0$ and $y=1$ is
- ✓
- B
$8 / 3 sq$. units
- C
$3 / 8$ sq. unit
- D
Answer(a) : We have, $x=3 y^2-9$
$
\Rightarrow 3 y^2=x+9
$
$\therefore \quad$ Required area
$
=\left|\int_0^1\left(3 y^2-9\right) d y\right|=\left|\left[y^3-9 y\right]_0^1\right|=|1-9|=8 \text { sq. units }
$ View full question & answer→MCQ 1911 Mark
The area bounded by the curve $y^2=x$, line $y=4$ and $y-$axis is
- A
$\frac{16}{3}$ sq. units
- ✓
$\frac{64}{3}$ sq. units
- C
$7 \sqrt{2}$ sq. units
- D
AnswerCorrect option: B. $\frac{64}{3}$ sq. units
We have, $y^2=x$ ， which is a parabola with vertex $(0,0)$ and line $y=4$
$\therefore \quad$ Required area
$=\int_0^4 y^2 d y$
$=\left[\frac{y^3}{3}\right]_0^4$
$=\frac{64}{3} \text { sq. units }$ View full question & answer→MCQ 1921 Mark
Which of these is equal to the area lying between the parabola $y^2=4 x$ and its latus rectum?
- A
$\frac{1}{3}$ sq. units
- B
$\frac{2}{3}$ sq. units
- C
$\frac{5}{3}$ sq. units
- ✓
$\frac{8}{3}$ sq. units
AnswerCorrect option: D. $\frac{8}{3}$ sq. units
(d) : We know that the area of region bounded by the parabola $y^2=4 a x$ and its latus rectum is $\frac{8}{3} a^2$ sq. units.
Here, $a=1$, therefore required area $=\frac{8}{3}$ sq. units View full question & answer→MCQ 1931 Mark
Area of the region bounded by the curve $y=x^2$ and the line $y=4$ is
- A
$\frac{11}{3}$ sq. units
- ✓
$\frac{32}{3}$ sq. units
- C
$\frac{43}{3}$ sq. units
- D
$\frac{47}{3} sq$. units
AnswerCorrect option: B. $\frac{32}{3}$ sq. units
(b) : We have, $y=x^2$
and $y=4$
$\therefore \quad$ Required area
$=2 \int_0^2\left(4-x^2\right) d x=\left[2\left(4 x-\frac{x^3}{3}\right)\right]_0^2=\frac{32}{3}$ sq. units View full question & answer→MCQ 1941 Mark
Find the area of the ellipse $\frac{x^2}{4^2}+\frac{y^2}{9^2}$.
- A
$16 \pi$ sq. unit
- B
$2 \pi$ sq. units
- ✓
$36 \pi$ sq. units
- D
$4 \pi$ sq. units
AnswerCorrect option: C. $36 \pi$ sq. units
(c) : Since, area of theellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi a b$ sq. units.
$\therefore \quad$ Required area $=\pi \times 4 \times 9=36 \pi$ sq. units.
View full question & answer→MCQ 1951 Mark
The area enclosed between the curve $y^2=4 x$ and the line $y=x$ is
- ✓
$\frac{8}{3}$ sq. units
- B
$\frac{4}{3}$ sq. units
- C
$\frac{2}{3}$ sq. units
- D
$\frac{1}{2}$ sq. units
AnswerCorrect option: A. $\frac{8}{3}$ sq. units
We have, $y^2=4 x ...(i)$
and $y=x ...(ii)$
$\therefore \quad$ Required area
$=\int_0^4(\sqrt{4 x}-x) d x$
$=\int_0^4\left(2 x^{1 / 2}-x\right) d x$
$=\left[2 \frac{x^{3 / 2}}{3 / 2}-\frac{x^2}{2}\right]_0^4$
$=\frac{4}{3}\left(4^{3 / 2}\right)-\frac{4^2}{2}$
$=\frac{32}{3}-8$
$=\frac{8}{3} \text { sq. units }$ View full question & answer→MCQ 1961 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
$\frac{45}{7}$ sq. units
- B
$\frac{25}{4}$ sq. units
- C
$\frac{5}{18}$ sq. units
- ✓
$\frac{9}{2}$ sq. units
AnswerCorrect option: D. $\frac{9}{2}$ sq. units
We have, $y=x^2+1 ...(i)$
and $x+y=3 ...(ii)$
Solving $(i)$ and $(ii),$ we get
$x^2+x-2=0$
$\Rightarrow x=-2,1$
$\therefore$ Required area
$=\int_{-2}^1\left\{3-x-\left(x^2+1\right)\right\} d x$
$=\left[2 x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-2}^1$
$=\left(2-\frac{1}{2}-\frac{1}{3}\right)-\left(-4-2+\frac{8}{3}\right)$
$=\frac{9}{2} \text { sq. units }$ View full question & answer→MCQ 1971 Mark
The area bounded by the curve $y=f(x)$, the $x$-axis and $x=1$ and $x=b$ is $(b-1) \sin (3 b+4)$. Which of the following can be $f(x)$ ?
AnswerCorrect option: C. $\sin (3 x+4)+3(x-1) \cdot \cos (3 x+4)$
(c) : Given, $\int_1^b f(x) d x=(b-1) \sin (3 b+4)$
Area function $=\int_1^x f(x) d x=(x-1) \sin (3 x+4)$
On differentiating, we get
$
f(x)=\sin (3 x+4)+3(x-1) \cdot \cos (3 x+4)
$
View full question & answer→MCQ 1981 Mark
The area enclosed by the curve $\frac{x^2}{25}+\frac{y^2}{9}=1$ is
- A
$10 \pi$ sq. units
- ✓
$15 \pi$ sq. units
- C
$5 \pi$ sq. units
- D
$4 \pi$ sq. units
AnswerCorrect option: B. $15 \pi$ sq. units
(b) : We have $\frac{x^2}{25}+\frac{y^2}{9}=1$, which is an ellipse Here, $a=5$ and $b=3$
Since, area of region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi a b$.
$\therefore \quad$ Required area $=\pi(5)(3)=15 \pi$ sq. units
View full question & answer→MCQ 1991 Mark
Which of these is equal to the area enclosed between the curve $x^2+y^2=16$ and the coordinate axes in the first quadrant?
- ✓
$4 \pi$ sq. units
- B
$3 \pi$ sq. units
- C
$2 \pi$ sq. units
- D
$\pi$ sq. units
AnswerCorrect option: A. $4 \pi$ sq. units
(a) : Given curve is a circle with centre $(0,0)$ and radius 4 .
$\therefore \quad$ Required area
$
=\int_0^4 \sqrt{16-x^2} d x
$
$=\left[\frac{x}{2} \sqrt{16-x^2}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_0^4=4 \pi$ sq. units View full question & answer→MCQ 2001 Mark
Area bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ is
- ✓
$6 \pi$ sq. units
- B
$3 \pi$ sq. units
- C
$12 \pi$ sq. units
- D
AnswerCorrect option: A. $6 \pi$ sq. units
(a) : Here $a^2=4$ and $b^2=9$.
Since, area of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi a b$ sq. units.
$\therefore \quad$ Required area $=\pi \times 2 \times 3=6 \pi$ sq. units.
View full question & answer→