RD Sharma Solutions Class 12 Mathematics Chapter 20 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 20 MCQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 03:18 PM IST

It is quite famous that the solutions of RD Sharma books are an efficient path for any student to follow to ace the subject of mathematics. The Class 12 RD Sharma chapter 20 exercise MCQ solution is one of the examples for it. The solution is constructive in every way out and is also exam-oriented, simple, and essential. The RD Sharma class 12th exercise MCQ can be used for self-practice and evaluation of marks.

## Areas of Bounded Regions Excercise:MCQ

Areas of Bounded Region exercise multiple choice question 1

1(b)
Hint:
Make the graph of 2kx and then apply integration
Given:
$y=2^{kx} \text { and } x=0, \quad x=2 \text { is }\frac{3}{log_{e}2}$
Explanation:

Area of the bounded region
$=\int_{0}^{2}2^{kx}dx$
\begin{aligned} &\frac{3}{log_{e}2}=\int_{0}^{2}2^{kx}dx \\ &\frac{3}{log_{e}2}=\left [ \frac{2^{kx}}{log_{e}2} \right ]_{0}^{2} \\ &3=\left [ 2^{kx} \right ]_{0}^{2} \\ &3=(2^{k})^{2}-(2^{k})^{0} \\ &3=2^{2k}-1 \end{aligned} \begin{aligned} &\left [ \because \int a^{x}dx=\frac{a^{x}}{log_{e}a} \right ] \\ &\left [ \because a^{0}=1 \right ] \end{aligned}
\begin{aligned} &4=2^{2k} \\ &2^{2}=2^{2k} \\ &2k=2 \\ &k=1 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 2

(c)
Hint:
Find the bounded region between the parabolas
Given:
y2 = 4x and x2 = 4y
Explanation:

Intersection point is
\begin{aligned} &y^{2}=4x \\ &=4(2\sqrt{y}) \qquad \qquad \qquad [\because x^{2}=4y] \\ &y^{2}=8\sqrt{y} \\ &y^{4}=64y \\ &y^{4}-64y=0 \\ &y(y^{3}-64)=0 \\ &y=0, \: \: y^{3}=64,\: \: y=4 \\ &\therefore x=4 \end{aligned}
Area between the parabolas
\begin{aligned} &=\int_{0}^{4}\left ( 2\sqrt{x}-\frac{x^{2}}{4} \right )dx \\ &=\frac{1}{4}\int_{0}^{4}(8\sqrt{x}-x^{2})dx \\ &=\frac{1}{4}\left [ \frac{8x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{3}}{3} \right ]_{0}^{4} \\ &=\frac{1}{4}\left [ \frac{16}{3}(4)^{\frac{3}{2}}-\frac{4^{3}}{3}-0+0 \right ] \\ &=\frac{1}{4}\left [ \frac{16}{3}(2^{2})^{\frac{3}{2}}-\frac{64}{3} \right ] \\ &=\frac{1}{4}\left [ \frac{128}{3}-\frac{64}{3} \right ]\\ &=\frac{16}{3}sq.units \end{aligned}

Areas of Bounded Regions exercise multiple choice question 3

(b)
Hint:
Construct the graph
Given:
y=logex and x - axis, the straight line x = e
Explanation:

Area bounded by the curve, x - axis and x = e
\begin{aligned} &=\int_{1}^{e}log_{e}x dx \\ &=[x.log_{e}x]_{1}^{e}-\int_{1}^{e}x.\frac{1}{x}dx \\ &=e\, log_{e}e-log_{e}1-\int_{1}^{e}dx \\ &=e(1)-0-[x]_{1}^{e} \qquad \qquad \qquad [\text { Integration by parts }]\\ &=e-[e-1] \\ &=e-e+1 \\ &=1sq.units \end{aligned}

Areas of Bounded Regions exercise multiple choice question 4

(b)
Hint:
Find the region between parabola and line
Given:
$y=2-x^{2},\quad x+y=0$
Explanation:

To find the point of intersection of
$y=2-x^{2} \text { and } x+y=0$
\begin{aligned} &x=-y \\ &y=2-x^{2} \\ &-x=2-x^{2} \\ &x^{2}-x-2=0 \\ &x^{2}-2x+x-2=0 \\ &x(x-2)+1(x-2)=0 \\ &(x+1)(x-2)=0 \\ &x=-1,2 \end{aligned}
Area of the region ABCD,
\begin{aligned} &A=\int_{-1}^{2}(2-x^{2}-(-x))dx \\ &=\int_{-1}^{2}(2-x^{2}+x)dx \\ &=\left [ 2x-\frac{x^{3}}{3}+\frac{x^{2}}{2} \right ]_{1}^{2} \\ &=\left [ 2(2)-\frac{(2)^{3}}{3} +\frac{(2)^{2}}{2}\right ]-\left [ 2(-1)-\frac{(-1)^{3}}{3} +\frac{(-1)^{2}}{2} \right ] \\ &=\left [ 4-\frac{8}{3}+2 \right ]-\left [ -2+\frac{1}{3}+\frac{1}{2} \right ] \\ &=6-\frac{8}{3}+2-\frac{1}{3}-\frac{1}{2} \\ &=5-\frac{1}{2} \\ &=\frac{9}{2}sq.units \end{aligned}

Areas of Bounded Regions exercise multiple choice question 5

$\frac{32}{3}(b)$
Hint:
Find the region between parabola and y - axis
Given:
x = 4 - y2 and y-axis (x=0)
Explanation:

Area bounded by region
\begin{aligned} &=\int_{-2}^{2}(4-y^{2})dy \\ &=\left [ 4y-\frac{y^{3}}{3} \right ]_{-2}^{2} \\ &=8-\frac{8}{3}+8-\frac{8}{3} \\ &=16-\frac{16}{3} \\ &=\frac{32}{3} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 6

(a)
Hint:
Find
$A_{n}>A_{n+1}\: or\: A_{n}
Given:
$y=(tan\, x)^{n}, x=0, y=0 \: and\: x=\frac{\pi }{4}$
Explanation:

\begin{aligned} &A_{n}=\int_{0}^{\frac{\pi }{4}}(tan\, x)^{n}dx \\ &A_{n-2}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, xdx \qquad\qquad\dots(i) \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n}\, xdx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: tan^{2}dx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: (sec^{2}x-1)dx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: sec^{2}x dx-\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, xdx \\ &A_{n}+A_{n-2}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: (sec^{2}x)dx \end{aligned}
Let,
\begin{aligned} &t=tan\: x \\ &dt=sec^{2}xdx\\ &x=0\: \: x=\frac{\pi }{4}\\ &t=0\: \: t=1\\ &A_{n}+A_{n-2}=\int_{0}^{1}t^{n-2}dt\\ &=\left [ \frac{t^{n-1}}{n-1} \right ]_{0}^{1}\\ &=\frac{1}{n-1} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 7

(c)
Hint:
Integration
Given:
$x^{2}+y^{2}-6x-4y+12\leq 0$
$y\leq x\: and\: x\leq \frac{5}{2}$
Explanation:
\begin{aligned} &x^{2}+y^{2}-6x-4y+12\leq 0\\ &x^{2}-6x+9-9+y^{2}-4y+4-4+12\leq 0\\ &(x-3)^{2}+(y-2)^{2}-1\leq 0\\ &(x-3)^{2}+(y-2)^{2}\leq 1\\ &(y-2)^{2}\leq 1-(x-3)^{2}\\ &(y-2)\leq\sqrt{1-(x-3)^{2}}\\ &y\leq2\pm \sqrt{1-(x-3)^{2}}\\ &\text{ As }x\leq \frac{5}{2},y=2- \sqrt{1-(x-3)^{2}} \end{aligned}
Area,
\begin{aligned} &=\int_{2}^{\frac{5}{2}}xdx-\int_{2}^{\frac{5}{2}}\left ( 2-\sqrt{1-(x-3)^{2}} \right )dx\\ &=\left [ \frac{x^{2}}{2} \right ]_{2}^{\frac{5}{2}}-\left [ 2x-\frac{(x-3)}{2}\sqrt{1-(x-3)^{2}}+\frac{1}{2}sin^{-1}(x-3) \right ]_{2}^{\frac{5}{2}}\\ &=\frac{25}{8}-2-\left [ 2\left ( \frac{1}{2} \right )-\frac{1}{2}\left ( \frac{-1}{2} \right )\times \left ( \frac{\sqrt{3}}{2} \right )+\frac{1}{2}sin^{-1}\left ( \frac{-1}{2} \right )-sin^{-1}(-1)\right ] \end{aligned}
\begin{aligned} &=\frac{9}{8}-1+\frac{\sqrt{3}}{8}+\frac{1}{2}\left ( \frac{-\pi }{6}+\frac{\pi }{2} \right )\\ &=\frac{1}{8}+\frac{\sqrt{3}}{8}-\frac{\pi }{6}\\ &=\frac{-\pi }{6}+\frac{1+\sqrt{3}}{8} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 8

(a)
Hint:
Integration
Given:
$y=log_{e}(x+e),x=log_{e}\left ( \frac{1}{y} \right ),x-axis$
Explanation:
\begin{aligned} &x-axis(y=0)\\ &0=log_{e}(x+e)\\ &x+e=e^{0}\\ &x+e=1\\ &x=1-e\\ &x=log_{e}\frac{1}{0}\\ &x=\infty \end{aligned}
Area,
\begin{aligned} &=\int_{1-e}^{0}\left [ log(x+e) \right ]dx+\int_{0}^{\infty }\left [ e^{-x} \right ]dx\\ &=\left [ (x+e)log(x+e) \right ]_{1-e}^{0}-\left [ \frac{(x+e)}{x+e} \right ]_{1-e}^{0}\lim_{k\rightarrow \infty }\left [ -e^{-x} \right ]_{0}^{k}\\ &=e[1-1]-[-1]+\lim_{k\rightarrow \infty }\left [ -e^{-x} \right ]_{0}^{k}\\ &=1+(1-0)\\ &=2 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 9

(c)
Hint:
Integration
Given:
(y-2)2 = x-1, the tangent to it at the point with ordinate 3 and x-axis.
Explanation:
(y-2)2 = x-1
Tangent,
\begin{aligned} &2(y-2)\frac{\mathrm{d} y}{\mathrm{d} x}=1\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2(y-2)} \end{aligned}
At y = 3
\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \end{aligned}
Therefore, the equation of tangent at (2,3)
\begin{aligned} &y-3=\frac{1}{2}(x-2)\\ &x-2y+4=0\\ &x=2y-4 \end{aligned}
The area of bounded region
\begin{aligned} &=\int_{0}^{3}\left [ (y-2)^{2}+1-(2y-4) \right ]dy\\ &=\int_{0}^{3}\left [ y^{2}-6y+9 \right ]dy\\ &=\left [ \frac{y^{3}}{3}-\frac{6y^{2}}{2}+3y \right ]_{0}^{3}\\ &=\frac{3^{3}}{3}\\ &=9 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 10

(a)
Hint:
Integration
Given:
$y=sin\, x, x=0\: \: and\: \: x=\pi ,x-axis$
Explanation:
\begin{aligned} &\int_{0}^{\pi }sin\, xdx\\ &=[-cos\, x]_{0}^{\pi }\\ &=-cos\, \pi +cos\, 0\\ &=1+1\\ &=2 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 11

(b)
Hint:
Integration
Given:
$y^{2}=4ax, x^{2}=4ay$
Explanation:
\begin{aligned} &\left ( \frac{x^{2}}{4a} \right )^{2}=4ax\\ &\frac{x^{4}}{16a^{2}}=4ax\\ &x^{4}-64a^{3}x=0\\ &x(x^{3}-64a^{3})=0\\ &x=0 \quad x=4a \end{aligned}
Area,
\begin{aligned} &=\int_{0}^{4a}\left [ 2\sqrt{ax}-\frac{x^{2}}{4a} \right ]dx\\ &=\left [ 2\sqrt{a}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{3}}{12a} \right ]_{0}^{4a}\\ &=\frac{4}{3}\sqrt{a}(4a)^{\frac{3}{2}}-\frac{(4a)^{3}}{12a}-0\\ &=\frac{4}{3}\sqrt{a}2a^{\frac{3}{2}}-\frac{16a^{2}}{3}\\ &=\frac{32a^{2}}{3}-\frac{16a^{2}}{3}\\ &=\frac{16a^{2}}{3} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 12

(b)
Hint:
Integration
Given:
y = x4 - 2x3 + x2 + 3 , x - axis and minima of y
Explanation:
Finding the minima of y
\begin{aligned} &y = x^{4} - 2x^{3} + x^{2} + 3 \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=4x^{3}-6x^{2}+2x \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=12x^{2}-12x+2 \\ \end{aligned}
Now,
\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &4x^{3}-6x^{2}+2x=0\\ &2x(2x^{2}-3x+1)=0\\ &x=0 \qquad 2x^{2}-3x+1=0\\ & \qquad \qquad (2x-1)(x-1)=0\\ &x=0,\: 1,\: \frac{1}{2} \end{aligned}
At x = 0
\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=2> 0 \end{aligned}
At x = 1
\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=12-12+2 =2> 0 \end{aligned}
At
\begin{aligned} &x=\frac{1}{2} \end{aligned}\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=-1 < 0 \end{aligned}

So,f(x) has minimum value at

\begin{aligned} &x=\frac{1}{2} \end{aligned}

The minimum value is y = 1

\begin{aligned} &1=x^{4}-2x^{3}+x^{2}+3\\ &x^{4}-2x^{3}+x^{2}+2=0\\ &x=1 \end{aligned}

\begin{aligned} &\int_{0}^{1} (x^{4}-2x^{3}+x^{2}+3)dx\\ &=\left [ \frac{x^{5}}{5}-2\left ( \frac{x^{4}}{x} \right )+\frac{x^{3}}{3}+3x \right ]_{0}^{1}\\ &=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3=0\\ &=\frac{91}{30} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 13

(b)
Hint:
Integration
Given:
$y^{2}=4ax, \text { latus rectum }=a , \text{ x-axis }(y=0)$
Explanation:

y = 0
x = 0
\begin{aligned} &=\int_{0}^{a}2\sqrt{ax}dx\\ &=2\sqrt{a}\int_{0}^{a}\sqrt{x}dx\\ &\text { Area }=2\sqrt{a}\left [ \frac{x^\frac{3}{2}}{\frac{3}{2}} \right ]_{0}^{a}\\ &=\frac{4}{3}\sqrt{a}\left [ a \right ]^{\frac{3}{2}}\\ &=\frac{4a^{2}}{3} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 14

(c)
Hint:
Integration
Given:
$[(x,y):x^{2}+y^{2}\leq 1\leq x+y]$
Explanation:

\begin{aligned} &x^{2}+y^{2}=1\\ &y^{2}=1-x^{2}\\ &y=\sqrt{1-x^{2}},\: 0< x< 1\\ &x+y=1\\ &y=1-x \end{aligned}
Area
\begin{aligned} &=\int_{0}^{1}(\sqrt{1-x^{2}})-(1-x)dx\\ &=\left [ \frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}sin^{-1}\frac{x}{2}-x+\frac{x^{2}}{2} \right ]_{0}^{1}\\ &=\frac{1}{2}sin^{-1}\left ( \frac{1}{2} \right )-1+\frac{1}{2}\\ &=\frac{1}{2}sin^{-1}\left ( \frac{1}{2} \right )-\frac{1}{2}\\ &=\frac{1}{2}\left ( \frac{\pi }{6}-1 \right ) \end{aligned}

Areas of Bounded Regions exercise 5 multiple choice question 15

(c)
Hint:
Integration
Given:
$y=2x^{2},\: \: y=x^{2}+4$
Explanation:
\begin{aligned} &2x^{2}=x^{2}+4\\ &x^{2}=4\\ &x=\pm 2\\ &\text { Area }=\int_{-2}^{2}(2x^{2}+4-2x^{2})dx\\ &=\int_{-2}^{2}(-x^{2}+4)dx \end{aligned}
\begin{aligned} &=\left [ \frac{-x^{3}}{3}+4x \right ]_{-2}^{2}\\ &=\frac{-8}{3}+8-\frac{8}{3}+8\\ &=16-\frac{16}{3} \\ &=\frac{48-16}{3}\\ &=\frac{32}{3}\text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 16

(d)
Hint:
Integration
Given:
$y=x^{2}+1,\: \: x+y=3$
Explanation:
Intersection points are
\begin{aligned} &x+y=3\\ &x+x^{2}+1=3\\ &x^{2}+x-2=0\\ &x^{2}+2x-x-2=0\\ &x(x+2)-1(x+2)=0\\ &x=1,\: \: x=-2 \end{aligned}
Area
\begin{aligned} &=\int_{-2}^{1}\left [ (3-x)-(x^{2}+1) \right ]dx\\ &=\int_{-2}^{1}\left [ 2-x-x^{2} \right ]dx\\ &=\left [ 2x-\frac{x^{2}}{2}-\frac{x^{3}}{3} \right ]_{-2}^{1}\\ &=2-\frac{1}{2}-\frac{1}{3}-\left [ -4-\frac{4}{2}+\frac{8}{3} \right ]\\ &=\frac{3}{2}-\frac{1}{3}+6-\frac{8}{3}\\ &=\frac{9}{2} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 17

(b)
Hint:
Integration
Given:
$\text { The ratio of } y=cos\, x \text { and } y=cos2x \text { from } x=0 \text { to }x=\frac{\pi }{3}$
Explanation:
\begin{aligned} &\text {Area of }cos\, x=A_{1}=\int_{0}^{\frac{\pi }{3}}cos\, xdx\\ &=\left [ sin\, x \right ]_{0}^{\frac{\pi }{3}}\\ &=sin\frac{\pi }{2}-sin\, 0\\ &=\frac{\sqrt{3}}{2} \end{aligned}
\begin{aligned} &\text {Area of }cos\, 2x=A_{2}=\int_{0}^{\frac{\pi }{3}}cos\, 2xdx\\ &=sin\frac{2\pi }{3}-sin\, 0\\ &=\frac{1}{2}sin\left ( \pi -\frac{\pi }{3} \right )-0\\ &=sin\frac{\pi }{3}\\ &=\frac{\frac{\sqrt{3}}{2}}{2}\\ &=\frac{\sqrt{3}}{4} \end{aligned}
\begin{aligned} &A_{1}:A_{2}=\frac{\left ( \frac{\sqrt{3}}{2} \right )}{\left ( \frac{\sqrt{3}}{4} \right )}=2:1 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 18

(d)
Hint:
Integration
Given:
$y=cos\, x,\: x-axis,\: x=0,\: x=2\pi$
Explanation:

\begin{aligned} &\int_{0}^{\frac{\pi }{2}}cos\, x+\int_{\frac{\pi }{2}}^{\frac{3\pi }{2}}(-cos\, x)dx+\int_{\frac{3\pi }{2}}^{2\pi }(cos\, x)dx\\ &=[sin\, x]_{0}^{\frac{\pi }{2}}+[-sin\, x]_{\frac{\pi }{2}}^{\frac{3\pi }{2}}+[sin\, x]_{\frac{3\pi }{2}}^{2\pi }\\ &=sin\frac{\pi }{2}-sin\, 0-\left [ sin\frac{3\pi }{2}-sin\frac{\pi }{2} \right ]+sin2\pi -sin\frac{3\pi }{2}\\ &=1+1+1+1\\ &=4 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 19

(a)
Hint:
Integration
Given:
$y^{2}=x,\: 2y=x$
Explanation:
Intersection point
\begin{aligned} &y^{2}=x\\ &y^{2}=2y\\ &y^{2}-2y=0\\ &y(y-2)=0\\ &y=0,\: \: y=2\\ &\Rightarrow x=4 \end{aligned}
Area
\begin{aligned} &=\int_{0}^{4}\left ( \sqrt{x}-\frac{x}{2} \right )dx\\ &=\left [ \frac{2}{3}x^{\frac{3}{2}} \right ]_{0}^{4}-\left [ \frac{x^{2}}{4} \right ]_{0}^{4}\\ &=\frac{2}{3}\times 8-4\\ &=\frac{4}{3} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 20

(c)
Hint:
Integration
Given:
$y=4x-x^{2} \text { and } x-axis (y=0)$
Explanation:
\begin{aligned} &y=0\\ &0=4x-x^{2}\\ &x^{2}-4x=0\\ &x(x-4)=0\\ &x=0,\: x=4 \end{aligned}
Area
\begin{aligned} &=\int_{0}^{4}(4x-x^{2})dx\\ &=\left [ \frac{4x^{2}}{2}-\frac{x^{3}}{3} \right ]_{0}^{4}\\ &=2(4)^{2}-\frac{(4)^{3}}{3}-0+0\\ &=32-\frac{64}{3}\\ &=\frac{96-64}{3}\\ &=\frac{32}{3}\text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 21

(b)
Hint:
Integration
Given:
$y^{2}(2a-x)=x^{3} \text { and } x=2a \text { above }x-axis$
Explanation:
\begin{aligned} &x=2a\\ &y^{2}(2a-x)=x^{3}\\ &y^{2}=\frac{x^{3}}{2a-x}\Rightarrow y=\sqrt{\frac{x^{3}}{2a-x}}\\ \end{aligned}
Let
\begin{aligned} &x=2a\: sin^{2}\theta \\ &dx=4a\: sin\: \theta\: cos\: \theta \end{aligned}
Area
\begin{aligned} &=\int_{0}^{2a}\sqrt{\frac{x^{3}}{2a-x}}dx\\ &=\int_{0}^{\frac{\pi }{2}}\sqrt{\frac{(2a)^{3}(sin^{2}\theta )^{3}}{2a-2asin^{2}\theta }}(4a\, sin\, \theta \: cos\, \theta\: d\theta ) \qquad \qquad [W\! hen\: x=0, \theta=0\: and\: x=2a,\theta =\frac{\pi }{2} ]\\ &=\int_{0}^{\frac{\pi }{2}}\sqrt{\frac{8a^{3}sin^{6}\theta }{2a\, cos^{2}\theta }}(4a\, sin\, \theta \: cos\, \theta\: d\theta ) \qquad \qquad [\because 1-sin^{2}\theta =cos^{2}\theta ] \end{aligned}
\begin{aligned} &=8a^{2}\int_{0}^{\frac{\pi }{2}}\sqrt{sin^{6}\theta }(sin\, \theta d\theta )\\ &=8a^{2}\int_{0}^{\frac{\pi }{2}}sin^{4}\theta d\theta \qquad \qquad [\because \int sin^{4}xdx=\frac{3x}{8}-\frac{1}{4}sin2x+\frac{1}{32}sin4x+c]\\ &=8a^{2}\left [ \frac{3\theta }{8}-\frac{1}{4}sin2\theta + \frac{1}{32}sin4\theta \right ]_{0}^{\frac{\pi }{2}}\\ &=8a^{2}\left [ \frac{3\left ( \frac{\pi }{2} \right )}{8}-\frac{1}{4}sin\pi +\frac{1}{32}sin2\pi \right ]-0\\ &=8a^{2}\times \frac{3\left ( \frac{\pi }{2} \right )}{8}\\ &=\frac{3}{2}\pi a^{2} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 22

(b)
Hint:
Integration
Given:
$x^{2}=4y,\: x=2 \text { and } x-axis (y=0)$
Explanation:
\begin{aligned} &y=0\\ &x^{2}=4(0)\\ &x=0\\ &\text { Area }=\int_{0}^{2}\frac{x^{2}}{4}dx\\ &=\left [ \frac{x^{3}}{12} \right ]_{0}^{2}\\ &=\frac{8}{12}\\ &=\frac{2}{3} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 23

(c)
Hint:
Integration
Given:
$\text { Area }=(b-1)sin(3b+4)\\ x=1,\: x=b$
Explanation:
\begin{aligned} &\text { Area }=\int_{1}^{b}f(x)dx\\ &(b-1)sin(3b+4)=\int_{1}^{b}f(x)dx \end{aligned}
Differentiate both sides with respect to b,
\begin{aligned} &f(b)=3(b-1)cos(3b+4)+sin(3b+4)\\ &\text { Put } b=x\\ &f(x)=3(x-1)cos(3x+4)+sin(3x+4) \end{aligned}

Areas of Bounded Regions exercise multiple choice question 24

(a)
Hint:
Integration
Given:
$y^{2}=8x,x^{2}=8y$
Explanation:
\begin{aligned} &y^{2}=8x\\ &x=\frac{y^{2}}{8}\\ &x^{2}=8y\\ &\left ( \frac{y^{2}}{8} \right )^{2}=8y\\ &\left ( \frac{y^{4}}{64} \right )=8y\\ &y^{4}-512y=0\\ &y(y^{3}-512)=0\\ &y=0,\: \: y^{3}=512\\ &y^{3}=8^{3}\\ &y=8 \end{aligned}
\begin{aligned} &\text { Area }=\int_{0}^{8}\left ( \sqrt{8x}-\frac{x^{2}}{8} \right )dx\\ &=\sqrt{8x}\int_{0}^{8}\sqrt{x}dx-\frac{1}{8}\int_{0}^{8}x^{2}dx\\ &=\sqrt{8}\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{8}-\frac{1}{8}\left [ \frac{x^{3}}{3} \right ]_{0}^{8}\\ &=\frac{2}{3}\sqrt{8}\left [ 8^{\frac{3}{2}}-0 \right ]-\frac{1}{24}\left [ 8^{3} -0\right ]\\ &=\frac{2\times 8^{2}}{3}-\frac{8^{2}}{3}\\ &=\frac{64}{3} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 25

(c)
Hint:
Integration
Given:
\begin{aligned} &y^{2}=8x, x-axis(y=0) \text { and latus rectum }\\ &y^{2}=4\times 2x,\: \text { latus rectum }=2\\ &\text {Area }=2\int_{0}^{2}\sqrt{8x}dx\\ &=4\sqrt{2}\int_{0}^{2}\sqrt{x}dx=4\sqrt{2}\left [ \frac{x^\frac{3}{2}}{\frac{3}{2}} \right ]_{0}^{2}\\ &=4\sqrt{2}\times \frac{2}{3}\left ( 2\frac{3}{2}-0 \right )\\ &=\frac{8\sqrt{2}}{3}\times 2\sqrt{2}\\ &=\frac{32}{3} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 26

(d)
Hint:
Integration
Given:
$y=x^{3},\: x-axis,\: x=-2,\: x=1$
Explanation:

Area OAB
\begin{aligned} &=\int_{0}^{1}x^{3}dx\\ &=\left [ \frac{x^{4}}{4} \right ]_{0}^{1}\\ &=\frac{1}{4} \end{aligned}
Area OCD
\begin{aligned} &=\int_{-2}^{0}x^{3}dx\\ &=\left [ \frac{x}{4} \right ]_{-2}^{0}\\ &=-4 \end{aligned}
As area is always positive
=4
Total area
\begin{aligned} &=4+\frac{1}{4}\\ &=\frac{17}{4} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 27

(c)
Hint:
Integration
Given:
$y=x\left | x \right |,\: x=-1,\: x=1$
Explanation:

$y=x\left | x \right |=\left\{\begin{matrix} x^{2} &x> 0 \\ -x^{2} & x< 0 \end{matrix}\right.$
Area of bounded region,
$=\int_{-1}^{1}x\left | x \right |dx$
Area AOB,
\begin{aligned} &=\int_{-1}^{0}-x^{2}dx\\ &=\left [ \frac{-x^{3}}{3} \right ]_{-1}^{0}\\ &=\frac{-1}{3} \end{aligned}
Area is always positive,
\begin{aligned} &=-\left ( \frac{-1}{3} \right )=\frac{1}{3} \end{aligned}
Area OCD,
\begin{aligned} &=\int_{0}^{1}x^{2}dx\\ &=\left [ \frac{x^{3}}{3} \right ]_{0}^{1}\\ &=\frac{1}{3} \end{aligned}
Total area
\begin{aligned} &=\frac{1}{3}+\frac{1}{3}\\ &=\frac{2}{3} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 28

(b)
Hint:
Integration
Given:
$y=cos\, x, y=sin\, x,0\leq x\leq \frac{\pi }{2}$
Explanation:

Intersection point,
\begin{aligned} &cos\, x=sin\, x\\ &-tan\, x=1\\ &x=\frac{\pi }{4}, 0\leq x\leq \frac{\pi }{2} \end{aligned}
Area AOB
\begin{aligned} &=\int_{0}^{\frac{\pi }{4}}(cos\, x-sin\, \, x)dx\\ &=\left [ sin\, x+cos\, x \right ]_{0}^{\frac{\pi }{4}}\\ &=sin\frac{\pi }{4}-sin\, 0+cos\frac{\pi }{4}-cos\, 0\\ &=\frac{1}{\sqrt{2}}+0+\frac{1}{\sqrt{2}}-1\\ &=-1+\frac{2}{\sqrt{2}}\\ &=-1+\sqrt{2}\\ &=\sqrt{2}-1 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 29

(c)
Hint:
Integration
Given:
$x^{2}+y^{2}=16,y^{2}=6x$
Explanation:

Area OABC
\begin{aligned} &=2 \text { Area }O\! AB\\ &=\left [ \text { Area }O\! AD+\text { Area }AB\! D \right ] \qquad \qquad \dots(i) \end{aligned}
Intersection points are
\begin{aligned} &x^{2}+y^{2}=16 \\ &x^{2}+6x=16 \qquad \qquad \qquad \left [ \because y^{2}=6x \right ]\\ &x^{2}+6x-16=0\\ &x^{2}+8x-2x-16=0\\ &x(x+8)-2(x+8)=0\\ &x-2=0 \qquad x+8=0\\ &x=2, x=-8 \end{aligned}
Not possible in first and third quadrant
x=2
From (i)
Area OABC,
\begin{aligned} &=2\left [ \int_{0}^{2}\sqrt{6x}dx+\int_{2}^{4}\sqrt{16-x^{2}}dx \right ]\\ &=2\left [ \sqrt{6}\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{2}+2\left [ \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}sin^{-1}\frac{x}{4} \right ]_{2}^{4}\\ &=\frac{4}{3}\sqrt{6}\left [ (2)^{\frac{3}{2}-1} \right ]+2\left [ \frac{4}{2}\sqrt{16-(4)^{2}}+8sin^{-1}\frac{4}{4} \right ]-\left [ \frac{2}{2}\sqrt{16-2^{2}}+8sin^{-1}\frac{2}{4} \right ] \end{aligned}
\begin{aligned} &=\frac{4\sqrt{6}}{3}\left ( 2\sqrt{2} \right )+2\left [ 0+2\times \frac{\pi }{2}-\sqrt{12}-8sin^{-1}\frac{1}{2} \right ]\\ &=\frac{16\sqrt{3}}{3}+2\left [ 4\pi -2\sqrt{3}-8\frac{\pi }{6} \right ] \\ &=\frac{16\sqrt{3}}{3}+8\pi -4\sqrt{3}-\frac{8\pi }{3}\\ &=\frac{4\sqrt{3}}{3}+\frac{16\pi }{3}\\ &=\frac{4}{3}\left ( 4\pi +\sqrt{3} \right ) \text { sq.units } \end{aligned}
Area of circle
\begin{aligned} &=\pi r^{2}\\ &=\pi (4)^{2}\\ &=16\pi \end{aligned}
Exterior area
\begin{aligned} &=16\pi -\frac{4}{3}\left ( 4\pi +\sqrt{3} \right )\\ &=16\pi -\frac{16}{3}\pi -\frac{4}{\sqrt{3}}\\ &=\frac{4}{3}(8\pi -\sqrt{3}) \text{sq.units} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 30

(b)
Hint:
Integration
Given:
x2 + y2 = 4 and x + y = 2
Explanation:

x2 + y2 = 4
r = 2
Area OAB,
\begin{aligned} &=\int_{0}^{2}\sqrt{4-x^{2}}-(2-x)dx\\ &=\int_{0}^{2}\sqrt{4-x^{2}}-2+xdx\\ &=\int_{0}^{2}\sqrt{4-x^{2}}dx+\int_{0}^{2}(x-2)dx\\ &=\left [ \frac{x}{2}\sqrt{4-x^{2}}+\frac{4}{2}sin^{-1}\frac{x}{2} \right ]_{0}^{2}+\left [ \frac{x^{2}}{2}-2x \right ]_{0}^{2}\\ &=\frac{2}{2}\sqrt{4-4}+\frac{4}{2}sin^{-1}\frac{2}{2}-0+2sin^{-1}0+\frac{4}{2}-4-0+0\\ &=2sin^{-1}1+2-4\\ &=2\times \frac{\pi }{2}-2\\ &=\pi -2 \end{aligned}

Areas of Bounded Regions exercise multiple choice question 31

(b)
Hint:
Integration
Given:
y2 = 4x and y = 2x
Explanation:

\begin{aligned} &y=2x\\ &y^{2}=4a\Rightarrow (2x)^{2}=4x\\ &4x^{2}=4x\\ &4x^{2}-4x=0\\ &4x(x-1)=0\\ &x=0,x=1 \end{aligned}
Area OAB
\begin{aligned} &=\int_{0}^{1}\left ( 2\sqrt{3}-2x \right )dx\\ &=\left [ \frac{2x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{2x^{2}}{2} \right ]_{0}^{1}\\ &=\left [ \frac{4}{3}x^{\frac{3}{2}}-x^{2} \right ]_{0}^{1}\\ &=\frac{4}{3}(1)^{\frac{3}{2}}-(1)^{2}-0+0\\ &=\frac{4}{3}-1\\ &=\frac{1}{3} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 32

(a)
Hint:
Integration
Given:
x2 + y2 = 4 and x = 0, x = 2
Explanation:

$x^{2}+y^{2}=4\Rightarrow y=\sqrt{4-x^{2}}$
Area OAB,
\begin{aligned} &=\int_{0}^{2}\sqrt{4-x^{2}}dx\\ &=\left [ \frac{x}{2}\sqrt{4-x^{2}}+\frac{4}{2}sin^{-1}\frac{x}{2} \right ]_{0}^{2}\\ &=\frac{2}{2}\sqrt{4-4}+2sin^{-1}\frac{2}{2}-0-2sin^{-1}0\\ &=2sin^{-1}1\\ &=2\left ( \frac{\pi }{2} \right )\\ &=\pi \end{aligned}

Areas of Bounded Regions exercise multiple choice question 33

(b)
Hint:
Integration
Given:
y2 =4x, y-axis (x = 0) and y=3
Explanation:

\begin{aligned} &x=0\\ &y^{2}=4(0)=0\\ &y=0 \text { and } y=3\\ &9=4x\\ &x=\frac{9}{4} \end{aligned}
Area CAB
\begin{aligned} &=\int_{0}^{\frac{9}{4}}(3-2\sqrt{x})dx\\ &=\left [ 3x-\frac{2x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{\frac{9}{4}}\\ &=\left [ 3x-\frac{4}{3}x^{\frac{3}{2}} \right ]_{0}^{\frac{9}{4}}\\ &=3\left ( \frac{9}{4} \right )-\frac{4}{3}\left ( \frac{9}{4} \right )^{\frac{3}{2}}-0+0\\ &=\frac{27}{4}-\frac{4}{3}\left ( \left ( \frac{3}{2} \right )^{2} \right )^{\frac{3}{2}}\\ &=\frac{27}{4}-\frac{4}{3}\times \frac{27}{8}\\ &=\frac{27}{4}-\frac{9}{2}\\ &=\frac{9}{4} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 34

(d)
Hint:
Area of the circle = 4 X (Area of circle in first quadrant)
Given:
x2 + y2 = 2
Explanation:

\begin{aligned} &x^{2}+y^{2}=2\\ &y^{2}=2-x^{2} \qquad \qquad \qquad \left [ r^{2}=2\Rightarrow r=\sqrt{2} \right ]\\ &y=\sqrt{2-x^{2}} \end{aligned}
Area of the circle = 4 X (Area of OAB)
\begin{aligned} &=4\int_{0}^{\sqrt{2}}\sqrt{2-x^{2}}dx\\ &=4\left [ \frac{x}{2}\sqrt{2-x^{2}}+\frac{2}{2}sin^{-1}\frac{x}{\sqrt{2}} \right ]_{0}^{\sqrt{2}}\\ &=4\left [ \frac{\sqrt{2}}{2}\sqrt{2-(\sqrt{2})^{2}}+1sin^{-1}\frac{\sqrt{2}}{\sqrt{2}}-\frac{0}{2}\sqrt{2-0}-sin^{-1}0 \right ]\\ &=4\left [ sin^{-1}1 \right ]\\ &=4\left ( \frac{\pi }{2} \right )\\ &=2\pi \text{ sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 35

(b)
Hint:
Area in all quadrants are equal. So, we do 4 X Area of one quadrant
Given:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Explanation:

Area of the ellipse
\begin{aligned} &=4\times \text { Area of OAB }\\ &=4\times \int_{0}^{a}\frac{b}{a}\sqrt{a^{2}-x^{2}}dx \qquad \qquad \qquad \dots (i)\\ &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ &\frac{y^{2}}{b^{2}}=1-\frac{x^{2}}{a^{2}}\\ &\frac{y^{2}}{b^{2}}=\frac{a^{2}-x^{2}}{a^{2}}\\ &y=\frac{b}{a}\sqrt{a^{2}-x^{2}} \end{aligned}
From (i)
\begin{aligned} &=\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx\\ &=\frac{4b}{a}\left [ \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a} \right ]_{0}^{a}\\ &=\frac{4b}{a} \left [ \frac{a}{2}\sqrt{a^{2}-a^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{a}{a}-\frac{0}{2}\sqrt{a^{2}-0}-\frac{a^{2}}{2}sin^{-1}\frac{0}{a} \right ]\\ &=\frac{4b}{a} \left [ 0+\frac{a^{2}}{2}sin^{-1}1-0-0 \right ]\\ &=\frac{4b}{a} \left [ \frac{a^{2}}{2}\left ( \frac{\pi }{2} \right ) \right ]\\ &=\pi ab \end{aligned}

Areas of Bounded Regions exercise multiple choice question 36

(b)
Hint:
Find intersection point then area
Given:
y = x2 and y = 16
Explanation:

\begin{aligned} &y=16\\ &16=x^{2}\\ &x=\pm 4 \end{aligned}
Area of the region
\begin{aligned} &=\int_{-4}^{4}16-x^{2}dx\\ &=\left [ 16x-\frac{x^{2}}{3} \right ]_{-4}^{4}\\ &=16(4+4)-\left [ \frac{(4)^{3}}{3}-\frac{(-4)^{3}}{3} \right ]\\ &=128-\left [ \frac{2\times 4\times 4\times 4}{3} \right ]\\ &=128-\frac{128}{3}\\ &=\frac{256}{3} \end{aligned}

Areas of Bounded Regions exercise multiple choice question 37

(b)
Hint:
$\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\left ( \frac{x}{a} \right )+c$
Given:
In first quadrant, x- axis (y = 0), y = x, x2 + y2 = 32
Explanation:

Intersection point,
\begin{aligned} &y = x\\ &x^{2}+x^{2}=32\\ &2x^{2}=32\\ &x^{2}=16\\ &x=\pm 4 \end{aligned}
x = -4 is not possible as we have to find in first quadrant and in first quadrant x > 0
\begin{aligned} &x=4\\ &y^{2}+x^{2}=32\\ &x^{2}=32\\ &x=4\sqrt{2} \end{aligned}
Now, y = x
Therefore, y = 4
Area of ABCD = Area of ABD + Area of BCD
\begin{aligned} &=\int_{0}^{4}xdx+\int_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}dx\\ &=\left [ \frac{x^{2}}{2} \right ]_{0}^{4}+\left [ \frac{x}{2}\sqrt{32-x^{2}}+\frac{1}{2}(32)sin^{-1}\frac{x}{4\sqrt{2}} \right ]_{4}^{4\sqrt{2}}\\ &=8+\frac{4\sqrt{2}}{2}\sqrt{32(4\sqrt{2})^{2}}+\frac{1}{2}(32)sin^{-1}\frac{4\sqrt{2}}{4\sqrt{2}}-\frac{4}{2}\sqrt{32-(4)^{2}}-16sin-1\frac{4}{4\sqrt{2}}\\ &=8+0+16sin^{-1}1-2\sqrt{32-16}-16sin^{-1}\frac{1}{\sqrt{2}} \end{aligned}
\begin{aligned} &=8+16\left ( \frac{\pi }{2} \right )-2(4)-16 \left ( \frac{\pi }{4} \right )\\ &=8+8\pi -8-4\pi \\ &=4\pi \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 38

(c)
Hint:
Put the limit y = -1,1
Given:
x = 2y + 3, y = 1, y = -1
Explanation:
Area of the bounded region
\begin{aligned} &=\int_{-1}^{1}(2y+3)dy\\ &=\left [ \frac{2y^{2}}{2}+3y \right ]_{-1}^{1}\\ &=\left [ y^{2}+3y \right ]_{-1}^{1}\\ &=1+3-(-1)^{2}-3(-1)\\ &=4-1+3\\ &=6 \text { sq. units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 39

(d)
Hint:
Find the intersection point
Given:
\begin{aligned} &x^{2}=4y\\ &x=4y-2 \qquad \qquad \qquad \dots (i) \end{aligned}
Explanation:
Intersection points are
\begin{aligned} &x^{2}=4y\\ &(4y-2)^{2}=4y\\ &16y^{2}+4-16y=4y\\ &16y^{2}-20y+4=0\\ &4y^{2}-5y+1=0\\ &4y^{2}-4y-y+1=0\\ &4y(y-1)-1(y-1)=0\\ &(y-1)(4y-1)=0\\ &y=1, y=\frac{1}{4} \\ &y=1\Rightarrow x=2\\ &y=\frac{1}{4}\Rightarrow x=4\left ( \frac{1}{4}\right )-2=-1 \qquad \text {[From (i)]}\end{aligned}
Area
\begin{aligned} &=\int_{-1}^{2}\left [ \left ( \frac{x+2}{4} \right )-\frac{x^{2}}{4} \right ]dx\\ &=\left [ \frac{x^{2}}{8}+\frac{2x}{4}-\frac{x^{2}}{12} \right ]_{-1}^{2}\\ &=\frac{1}{4}\left [ \left ( 2+4-\frac{8}{3} \right )-\left ( \frac{1}{2}-2+\frac{1}{3} \right ) \right ]\\ &=\frac{1}{4}\left [ \frac{10}{3}-\left ( \frac{-7}{6} \right ) \right ]\\ &=\frac{1}{4}\left ( \frac{9}{2} \right )\\ &=\frac{9}{8} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 40

(a)
Hint:
$\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\left ( \frac{x}{a} \right )+c$
Given:
$y= \sqrt{16-x^{2}}, x-axis(y=0)$
Explanation:
\begin{aligned} &y=0\\ &0=\sqrt{16-x^{2}}\\ &16-x^{2}=0\\ &x^{2}=16\\ &x=\pm 4\\ \end{aligned}
Area bounded by the curve
\begin{aligned} &=\int_{-4}^{4}\sqrt{16-x^{2}}dx\\ &=\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{4^{2}}{2}sin^{-1}\frac{x}{4} \right ]_{-4}^{4}\\ &=\left [ \frac{4}{2}\sqrt{16-4^{2}}+\frac{4^{2}}{2}sin^{-1}\frac{4}{4} \right ]-\left [ \frac{-4}{2}\sqrt{16-(-4)^{2}}+\frac{4^{2}}{2}sin^{-1}\left ( \frac{-4}{4} \right ) \right ] \qquad \qquad \left [ \because sin^{-1}1=\frac{\pi }{2} \right ]\\ &=0+8sin^{-1}1-0+8sin^{-1}1\\ &=8\times 2\times \frac{\pi }{2}\\ &=8\pi \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise multiple choice question 41

(a)
Hint:
Find the graph between y2 = x and 2y = x
Given:
y2 = x and 2y = x
Explanation:

Intersection point,
\begin{aligned} &y=\frac{x}{2}\\ &y^{2}=x\\ &\left ( \frac{x}{2} \right )^{2}=x\Rightarrow \frac{x^{2}}{4}=x\\ &x^{2}-4x=0\\ &x(x-4)=0\\ &x=0,x=4 \end{aligned}
Area of the bounded region ABC
\begin{aligned} &=\int_{0}^{4}\left ( \sqrt{x}-\frac{x}{2} \right )dx\\ &=\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{2}}{4} \right ]_{0}^{4}\\ &=\frac{2}{3}(4)^\frac{3}{2}-\frac{(4)^{2}}{4}\\ &=\frac{4}{3} \text { sq.units } \end{aligned}

The RD Sharma class 12 solution of Area of bounded region exercise MCQ is used by thousands of students and teachers for the practical knowledge of maths. The RD Sharma class 12th exercise MCQ consists of a total of 41 questions that covers up all the essential concepts of the chapter that is mentioned below:

• Algorithm to find Area using vertical and horizontal stripes

• Method to find the Area between two curves

• The Area between two curves using vertical and horizontal stripes

• Area of the region

• Area of the region bounded by the curve

• Area of the region bounded by a parabola

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