RD Sharma Class 12 Solutions Chapter20 MCQ Areas Of Bounded Region - Other Exercise

• Chapter 20 - Areas Of Bounded Region - Ex-20.1

• Chapter 20 - Areas Of Bounded Region - Ex-20.2

• Chapter 20- Areas Of Bounded Region - Ex-20.3

• Chapter 20- Areas Of Bounded Region - Ex-20.4

• Chapter 20 - Areas Of Bounded Region - Ex-FBQ

Areas of Bounded Regions Excercise:MCQ

Areas of Bounded Region exercise multiple choice question 1

Answer:
1(b)
Hint:
Make the graph of 2^kx 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

Answer:
(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

Answer:
(b)
Hint:
Construct the graph
Given:
y=log_ex 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

Answer:
(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

Answer:
$\frac{32}{3}(b)$
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

Answer:
(a)
Hint:
Find
$A_{n}>A_{n+1}\: or\: A_{n}<A_{n+1}$
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 13

Answer:
(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

Answer:
(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 multiple choice question 18

Answer:
(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 26

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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

Answer:
(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 41

Answer:
(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
$\begin{aligned} &=\int_{0}^{4}\left ( \sqrt{x}-\frac{x}{2} \right )dx\\ &=\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{2}}{4} \right ]_{0}^{4}\\ &=\frac{2}{3}(4)^\frac{3}{2}-\frac{(4)^{2}}{4}\\ &=\frac{4}{3} \text { sq.units } \end{aligned}$

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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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RD Sharma Chapter-wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable