RD Sharma Class 12 Exercise 20.1 Areas of bounded region Solutions Maths - Download PDF Free Online

Areas of Bounded Regions Excercise: 20.1

Areas of Bounded Region exercise 20.1 question 2

Answer: $\frac{17}{2}sq\cdot units$
Hint: Use $- x+ y = 1$
Given: Using integration, find area of region bounded by line $y-1=x$ the $x-axis$ and ordinate$x=-2$ and $x=3$
Solution:

Here$y-1=x$is equation of line

We can write

_{$\begin{aligned} &y=x+1 \\\\ &-x+y=1 \end{aligned}$}

We get_{$\frac{-x}{1}+\frac{y}{1}=1$}

Consider $AB$ as line intersecting the $x-axis$ at point _$C(-1,0)$

So required area_{$=\text { Area of } C D A C+\text { Area of } C B E C$}

_{$=\int_{-1}^{3} y d x+\int_{-2}^{-1}-(y) d x$}

Substituting the value of_$y$,

_{$\begin{aligned} &=\int_{-1}^{3}(x+1) d x+\int_{-2}^{-1}-(x+1) d x \\\\ &=\left[\frac{x^{2}}{2}+x\right]_{-1}^{3}-\left[\frac{x^{2}}{2}+x\right]_{-2}^{-1} \end{aligned}$ $\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]$}

Substituting the value of_$x$ ,

$\begin{aligned} &=\left(\frac{9}{2}+3\right)-\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}-1\right)-(2-2) \\\\ &=\left(\frac{15}{2}+\frac{1}{2}\right)-\left(\frac{-1}{2}\right) \\\\ &=8+\frac{1}{2} \\ &=\frac{17}{2} s q \cdot \text { units } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 3

Answer:
$\frac{8}{3}a^{2}\; sq\cdot units$
Hint:
Find the shaded area
Given:
Find area of region bounded by parabola $y^{2}=4 a x$ and the line $x=a$
Solution:

We have
$x=a$ … (i)
$y^{2}=4 a x$ … (ii)
Required area $=$ shaded region $OBAO$
$\left[\because y^{2}=4 a x \Rightarrow y=\sqrt{4 a x}\right]$
$=2(\text { Shaded region } \mathrm{OBCO}) \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]$
$\begin{aligned} =& 2 \int_{0}^{a} \sqrt{4 a x} d x \\\\ &=2 \sqrt{4 a} \int_{0}^{a}(x)^{\frac{1}{2}} d x \\\\ &=2 \sqrt{4 a}\left[\frac{(x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{a} \end{aligned}$
$\begin{aligned} &=2 \sqrt{4 a}\left[\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{a} \\\\ &=2 \sqrt{4 a} \times \frac{2}{3}\left[(x)^{\frac{3}{2}}\right]_{0}^{a} \end{aligned}$
$\begin{aligned} &=\frac{4 \sqrt{4 a}}{3}\left[(a)^{\frac{3}{2}}-(0)^{\frac{3}{2}}\right] \\\\ &=\frac{4 \sqrt{4 a}}{3} \times(\sqrt{a})^{3} \\\\ &=\frac{4 \sqrt{4 a}}{3} \times(\sqrt{a} \times \sqrt{a} \times \sqrt{a}) \end{aligned}$
$\begin{aligned} &=\frac{4 \sqrt{4 a}}{3} \times a \sqrt{a} \\\\ &=\frac{8 a^{2}}{3} \text { sq.unit } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 4

Answer:
$\frac{32}{3}sq\cdot units$
Hint:
Solve area of integration.
Given:
Find area lying above the $x-axis$ and under parabola $y=4x-x^{2}$
Solution:

$\begin{aligned} &A=\int_{0}^{4}|y| d x\\\\ &A=\int_{0}^{4} y d x \end{aligned}$ $\text { [As } y>0 \text { for } 0,0 \leq x \leq 4|y|=4]$
$\begin{aligned} &A=\int_{0}^{4}\left(4 x-x^{2}\right) d x \\\\ &A=\left[\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4} \\\\ &A=32-\frac{64}{3} \\ &A=\frac{32}{3} s q \cdot u n i t \end{aligned}$

Areas of Bounded Region exercise 20.1 question 5

Answer:
$8\sqrt{3}\; sq\cdot units$
Hint:
$A=2 \times \text { Area of } O A B O$
Given:
Draw rough sketch to indicate the region bounded between curve $y^{2}=4 x$ and line $x=3$ .Also find the area of region.
Solution:

$y^{2}=4 x$ Represent parabola with vertex at $x=3$ line parallel to$y-axis$.
Since $y^{2}=4 x$ is symmetrical axis
Area of corresponding rectangle $=\left ( y \right )dx$
$\begin{aligned} &A=2 \times \text { Area of } O A B O \\\\ &A=2 \int_{0}^{3}|y| d x \\\\ &A=2 \int_{0}^{3} y d x \end{aligned}$
$\begin{aligned} &A=2 \int_{0}^{3} \sqrt{4 x} d x \\\\ &A=2 \times 2 \int_{0}^{3} \sqrt{x} d x \\\\ &A=4 \int_{0}^{3} \sqrt{x} d x \end{aligned}$
$\begin{aligned} &A=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{3} \\\\ &=\frac{8}{3}\left[(x)^{\frac{3}{2}}\right]_{0}^{3} \\\\ &=\frac{8}{3} \times 3 \sqrt{3} \\\\ &=8 \sqrt{3} s q \cdot \text { unit } \end{aligned}$ $\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]$

Areas of Bounded Region exercise 20.1 question 6

Answer:
$\frac{16}{3}sq\cdot units$
Hint:
$y=4-x^{2}, x=0, x=2$
Given:
Make rough sketch of graph of function $y=4-x^{2}$ ,$0\leq x\leq 2$ and determine the area enclosed by curve and $x-axis$ ,line $x=0$ and $x=2$ .
Solution:

$y=4-x^{2}$ , $0\leq x\leq 2$ represent half parabola with vertex $\left ( 2,0 \right )$
$x=2$ represent a line parallel to $y-axis$ and cutting axis at $\left ( 2,0 \right )$
Area required $= y dx$
$\begin{aligned} &A=\text { Area of } O A B O \\\\ &A=\int_{0}^{2}|y| d x \\\\ &A=\int_{0}^{2} y d x \end{aligned}$
$\begin{aligned} &A=\int_{0}^{2}\left(4-x^{2}\right) d x \\\\ &A=\left[4 x-\frac{x^{3}}{3}\right]_{0}^{2} \end{aligned}$
$\begin{aligned} &A=8-\frac{8}{3} \\\\ &A=\frac{16}{3} s q \cdot u n i t \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \end{aligned}$

Areas of Bounded Region exercise 20.1 question 7

Answer:
$\frac{2}{3}\left(5^{\frac{3}{2}}-1\right)$
Hint:
Where $y=\sqrt{x+1} \text { in }[0,4]$
Given:
Sketch the graph of _{$y=\sqrt{x+1}$}in_{$\left [ 0,4 \right ]$}and determine the area of region enclosed by curve, the_$x-axis$and line _$x=0,x=4$
Solution:

$\text { Area of } O A B C O=\int_{0}^{4}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]$
$\begin{aligned} &A=\int_{0}^{4} y d x \\\\ &A=\int_{0}^{4} \sqrt{x+1} d x \\\\ &A=\int_{0}^{4}(x+1)^{\frac{1}{2}} d x \end{aligned}$
$\begin{aligned} &A=\left[\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} \\\\ &A=\frac{2}{3}\left(5^{\frac{3}{2}}-1\right) s q \cdot \text { unit } \end{aligned}$ $\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]$

Areas of Bounded Region exercise 20.1 question 8

Answer:
$\frac{56}{9}sq\cdot units$
Hint:
$y=\sqrt{6 x+4} \quad \text { at } x=0, x=2$
Given:
Find area under curve $y=\sqrt{6 x+4}$ above $x-axis$ form $x=0$ to $x=2$ . Draw sketch of curve also.
Solution:

$y=\sqrt{6 x+4}$ represent a parabola with vertex $v=\left(\frac{-2}{3}, 0\right)$ and symmetric about$x-axis$ where $x=0$ is $y-axis$
The rectangle move from $x=0$ to $x=2$
Consider,
$\begin{aligned} &\text { Area } O A B C=\int_{0}^{2}|y| d x \\\\ &A=\int_{0}^{2} \sqrt{6 x+4} d x \\\\ &A=\int_{0}^{2}(6 x+4)^{\frac{1}{2}} d x \end{aligned}$
$\begin{aligned} &A=\frac{1}{6}\left[\frac{(6 x+4)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\\\ &A=\frac{2}{18}\left[(16)^{\frac{3}{2}}-(4)^{\frac{3}{2}}\right] \end{aligned}$
$\begin{aligned} &A=\frac{2}{18}\left[4^{3}-2^{3}\right] \\\\ &A=\frac{2}{18}[64-8] \\\\ &A=\frac{2}{18}[56] \\\\ &A=\frac{56}{9} s q \cdot \text { unit } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 9

Answer:
$\frac{4}{3}sq\cdot unit$
Hint:
$\text { Area of } A B C A=2 \times \text { Area of } A B D A$
Given:
Draw the rough sketch of $v^{2}+1=x, x \leq 2$ .Find area enclosed by curve and line $x=2$
Solution:

$\begin{aligned} &y^{2}+1=x, x \leq 2\\\\ &y^{2}+1=x \Rightarrow y=\sqrt{x-1}\\\\ &\text { Area of } A B C A=2 \times \text { Area of } A B D A\\\\ &A=2 \int_{1}^{2}|y| d x \end{aligned}$
$\begin{aligned} &A=2 \int_{1}^{2} y d x \quad[y>0 \Rightarrow|y|=y] \\\\ &A=2 \int_{1}^{2} \sqrt{x-1} d x \end{aligned}$
$\begin{aligned} &A=2\left[\frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{2} \\\\ &A=\frac{4}{3}\left[1^{\frac{3}{2}}-0\right] \\\\ &A=\frac{4}{3} s q \cdot \text { unit } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 10

Answer:
$3\pi \; sq\cdot unit$
Hint:
$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
Given:
Draw rough sketch of graph of curve$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$and evaluate area of region under curve and above$x-axis$
Solution:

Since given equation$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
All power of $x$ and $y$ are even
A = Area of enclosed curve along $x-axis$
$\begin{aligned} &=2 \int_{0}^{2}|y| \mathrm{dx} \\\\ &A=2 \int_{0}^{2} y \mathrm{dx} \\\\ &A=2 \int_{0}^{2} \frac{3}{2} \sqrt{4-x^{2}} d x \end{aligned}$
$\begin{aligned} &A=3 \int_{0}^{2} \sqrt{4-x^{2}} d x \\\\ &A=3\left[\frac{1}{2} x \sqrt{4-x^{2}}+\frac{1}{2} 4 \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \end{aligned}$
$\begin{aligned} &A=3\left[0+\frac{1}{2} \times 4 \sin ^{-1} 1\right] \\\\ &A=3 \times \frac{1}{2} \times 4 \times \frac{\pi}{2} \\\\ &A=3 \pi \text { sq.units } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 11

Answer:
$6\pi \; sq\cdot unit$
Hint:
$9 x^{2}+4 y^{2}=36$
Given:
Sketch the region $(x, y): 9 x^{2}+4 y^{2}=36$ and find the area of region enclosed by using integration.
Solution:

We have,
$9 x^{2}+4 y^{2}=36$ .........(i)
$\begin{aligned} &4 y^{2}=36-9 x^{2} \\\\ &y^{2}=\frac{9}{4}\left(4-x^{2}\right) \\\\ &y=\frac{3}{2} \sqrt{4-x^{2}} \end{aligned}$ ...........(ii)
From (i), we get
$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
Since $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ all power of $x$ and $y$ are equal
$\begin{aligned} &A=\text { Area of enclosed curve }=4 \int_{0}^{2}|y| d x \\\\ &A=4 \int_{0}^{2} \frac{3}{2} \sqrt{4-x^{2}} d x \\\\ &A=4 \times \frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} d x \end{aligned}$
$\begin{aligned} &A=6 \int_{0}^{2} \sqrt{2^{2}-x^{2}} d x \\\\ &A=6\left[\frac{x}{2} \sqrt{2^{2}-x^{2}}+\frac{1}{2} 2^{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \\\\ &A=6\left[0+\frac{1}{2} 4 \sin ^{-1} 1\right] \end{aligned}$
$\begin{aligned} &A=6\left[\frac{1}{2} \times 4\left(\frac{\pi}{2}\right)\right] \\\\ &A=6 \pi \mathrm{sq} \cdot \text { unit } \\\\ &{\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right] \quad \text { and }\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]} \end{aligned}$

Areas of Bounded Region exercise 20.1 question 12

Answer:
$\frac{\pi}{2} \; sq\cdot unit$
Hint:
$\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]$
Given:
Draw a rough sketch of graph of function $y=2 \sqrt{1-x^{2}}, x \in[0,1]$ and evaluate the area enclosed between the curve and $x-axis$
Solution:

We have
$\begin{aligned} &y=2 \sqrt{1-x^{2}} \\\\ &\frac{y}{2}=\sqrt{1-x^{2}} \\\\ &\frac{y^{2}}{4}=1-x^{2} \end{aligned}$
$\begin{aligned} x^{2}+\frac{y^{2}}{4} &=1 \\\\ \frac{x^{2}}{1}+\frac{y^{2}}{4} &=1 \end{aligned}$
Since given equation $\frac{x^{2}}{1}+\frac{y^{2}}{4}=1$
$\begin{aligned} &A=\int_{0}^{1} 2 \sqrt{1-x^{2}} d x \\\\ &A=2 \int_{0}^{1} \sqrt{1-x^{2}} d x \\\\ &A=2\left[\frac{1}{2} x \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{0}^{1} \end{aligned}$
$\begin{aligned} &A=2\left[\frac{1}{2} \sin ^{-1} 1\right] \\\\ &A=\sin ^{-1} 1 \\\\ &A=\frac{\pi}{2} s q \cdot \text { unit } \end{aligned}$ $\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]$

Areas of Bounded Region exercise 20.1 question 13

Answer:
$\frac{\pi a^{2}}{4} s q \cdot \text { unit }$
Hint:
$y=\sqrt{a^{2}-x^{2}} \text { at line } x=0, x=a$
Given: Determine the area under curve $y=\sqrt{a^{2}-x^{2}}$ include between line $x=0$ and $x=a$
Solution:

$\begin{aligned} &y=\sqrt{a^{2}-x^{2}} \\\\ &y^{2}=a^{2}-x^{2} \\\\ &x^{2}+y^{2}=a^{2} \end{aligned}$
Required area $=$ area of shaded region
$\begin{aligned} &=\int_{0}^{a} y d x \\\\ &=\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x \\\\ &=\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a} \end{aligned}$
$\begin{aligned} &=\left[\frac{a}{2} \sqrt{a^{2}-a^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{a}{a}\right]-\left[\frac{0}{2} \sqrt{a^{2}-0}+\frac{a^{2}}{2} \sin ^{-1} \frac{0}{a}\right] \\\\ &=\frac{a^{2}}{2} \sin ^{-1} 1 \\\\ &=\frac{a^{2}}{2} \times \frac{\pi}{2} \\\\ \end{aligned}$
$=\frac{\pi a^{2}}{4} s q \cdot u n i t$

Areas of Bounded Region exercise 20.1 question 15

Answer:
$\pi a^{2}\; s q \cdot \text { unit }$
Hint:
Use definite integral
Given:
Using definite integral, find area of circle $x^{2}+y^{2}=a^{2}$
Solution:

$x^{2}+y^{2}=a^{2}$
Centre$=\left ( 0,0 \right )$
Radius$=a$
Hence, $OA=OB=Radius=a$
$A=(a,0),B=(0,a)$
Area of circle$=4\times$ area of region $OBAO$
$=4 \int_{0}^{a} y d x$
We know that
$\begin{aligned} &x^{2}+y^{2}=a^{2} \\\\ &y^{2}=a^{2}-x^{2} \\\\ &y=\pm \sqrt{a^{2}-x^{2}} \end{aligned}$
Since $AOBA$ lies in first quadrant, value of $y$ is positive
$y=\sqrt{a^{2}-x^{2}}$
Now,
Area of circle $=4 \int_{0}^{a} \sqrt{a^{2}-x^{2}} d x \quad\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]$
$\begin{aligned} &=4\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a} \\\\ &=4\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{a}{a}\right]-\left[\frac{0}{2} \sqrt{a^{2}-0}+\frac{0^{2}}{2} \sin ^{-1} \frac{0}{a}\right] \end{aligned}$
$\begin{aligned} &=4\left[0+\frac{a^{2}}{2} \sin ^{-1} 1\right]_{0}^{a} \\\\ &=4 \times \frac{a^{2}}{2} \times \frac{\pi}{2} \\\\ &=\pi a^{2} \text { sq.units } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 16

Answer:
$\frac{27}{2} s q \cdot \text { units }$
Hint:
$y=1+|x+1|, x=-2, x=3, y=0$
Given:
Using integration, find area of region bounded by following curve making a rough sketch
$y=1+|x+1|, x=-2, x=3, y=0$
Solution:

We have
$y=1+|x+1|$ Intersect $x=-2$ at $\left ( -2,2 \right )$ and $x=3$ at $(3,5)$
$y=0$ is $x-axis$

$y=1+|x+1|$

$\left\{\begin{array}{cc} 1-(x+1) & x \leq-1 \\ 1+(x+1) & x \geq 1 \\ -x & x \leq-1 \\ x+2 & x \geq 1 \end{array}\right.$
Let required area be $A$ since limits on $x$ are given we use horizontal strip to find area
$\begin{aligned} &A=\int_{-2}^{3}|y| d x \\\\ &A=\int_{-2}^{-1}|y| d x+\int_{-1}^{3}|y| d x \\\\ &A=\int_{-2}^{-1}-x d x+\int_{-1}^{3}(x+2) d x \end{aligned}$
$\begin{aligned} &A=-\left[\frac{x^{2}}{2}\right]_{-2}^{-1}+\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{3} \\\\ &A=-\left[\frac{1}{2}-\frac{4}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \\\\ &A=\frac{3}{2}+8+\frac{8}{2} \end{aligned}$
$\begin{aligned} &A=\frac{3}{2}+8+4 \\\\ &A=\frac{27}{2} s q \cdot u n i t \end{aligned} \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]$

Areas of Bounded Region exercise 20.1 question 17

Answer:
$\frac{9}{2}\; s q \cdot unit$
Hint:
Break the limit and find the value of integral.
$\int_{0}^{1}-(x-5) d x$
Given:
Sketch the graph $y=\left | x-5 \right |$
Evaluate$\int_{0}^{1}|x-5| d x$. What the value of integral represents on graph
Solution:

$\begin{aligned} &\int_{0}^{1}|x-5| d x=\int_{0}^{1}-(x-5) d x \\\\ &=\left[\frac{-x^{2}}{2}+5 x\right]_{0}^{1} \\\\ &=\left[\frac{-1}{2}+5\right] \end{aligned}$
$=\frac{9}{2}\; s q \cdot unit$ $\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]$

Areas of Bounded Region exercise 20.1 question 18

Answer:
$9\; s q \cdot \text { units }$
Hint:
Break the limit and find integral
Given:
_{$y=\left | x+3 \right |$} and evaluate _{$\int_{-6}^{0}|x+3| d x$} What represent of graph
Solution:

Let's draw graph,_{$y=\left | x+3 \right |$
$y=|x+3|=\left\{\begin{array}{cc} x+3 & \text { for } x+3 \geq 0 \\ -(x+3) & \text { for } x+3<0 \end{array}\right.$
$\left\{\begin{array}{c} x+3 \quad \text { for } x \geq-3 \\ -(x+3) \text { for } x+3<-3 \end{array}\right.$
$\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]$}
Required area
_{$\begin{aligned} &=\int_{-6}^{0}|x+3| d x \\\\ &=\int_{-6}^{-3}|x+3| d x-=\int_{-3}^{0}|x+3| d x \\\\ &=\int_{-6}^{-3}-(x+3) d x-\int_{-3}^{0}(x+3) d x \end{aligned}$
$\begin{aligned} &=\left[\frac{-x^{2}}{2}-3 x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+3 x\right]_{-3}^{0} \\\\ &=\left[\frac{-(-3)^{2}}{2}-3(-3)\right]_{-6}^{-3}-\left[\frac{-(-6)^{2}}{2}-3(-6)\right]+\left[\frac{0^{2}}{2}+3(0)\right]-\left[\frac{(-3)^{2}}{2}+3(-3)\right] \end{aligned}$
$\begin{aligned} &=\frac{-9}{2}-(-9)-\frac{-36}{2}-(-18)+\left[0+\left(\frac{-9}{2}+9\right)\right] \\\\ &=\frac{-9}{2}+9+0-\frac{9}{2}+9 \\\\ &=-9+18 \\\\ &=9 \text { sq.units } \end{aligned}$}

Areas of Bounded Region exercise 20.1 question 19

Answer:
$9\; s q \cdot { units }$
Hint:
Find integral $\int_{-4}^{2}|x+1| d x$
Given:
$y=|x+1|$ Evaluate $\int_{-4}^{2}|x+1|$ what does the value of integral represented on graph
Solution:

We have,
$y=|x+1|$ intersect $x = -4$ and $x = 2$ at $( -4,3 )$ and $( 2,3 ).$
Now,
$y=|x+1|$
$=\left\{\begin{array}{c} (x+1) \text { for all } x>-1 \\ -(x+1) \text { for all } x<-1 \end{array}\right.$
Integral represents the area enclosed between $x = -4$ and $x = 2.$
$\begin{aligned} A &=\int_{-4}^{2}|y| d x \\\\ A &=\int_{-4}^{-1}|y| d x+\int_{-1}^{2}|y| d x \\\\ A &=\int_{-4}^{-1}-(x+1) d x+\int_{-1}^{2}(x+1) \end{aligned}$
$\begin{aligned} &A=-\left[\frac{x^{2}}{2}+x\right]_{-4}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{2} \\\\ &A=-\left[\frac{1}{2}-1-\frac{16}{2}+4\right]+\left[\frac{4}{2}+2-\frac{1}{2}+1\right] \end{aligned}$
$\begin{aligned} &A=-\left[3-\frac{15}{2}\right]+\left[5-\frac{1}{2}\right] \\\\ &A=-3+\frac{15}{2}+5-\frac{1}{2} \\\\ &A=\frac{-6+15+10-1}{2} \end{aligned}$
$A=9\; s q \cdot { units }$

Areas of Bounded Region exercise 20.1 question 20

Answer:
$3+16\log 2\; s q \cdot { unit }$
Hint:
Use the concept of definite integrals.
Given:
Find area of region bounded by curve_{$x y-3 x-2 y-10=0, x-a x i s$}and line _{$x=3, x=4$}
Solution:
Given curve,
$\begin{aligned} &x y-3 x-2 y-10=0 \\\\ &x y-2 y=3 x+10 \\\\ &y(x-2)=3 x+10 \\\\ &y=\frac{3 x+10}{x-2} \end{aligned}$

Area of bounded curve $x y-3 x-2 y-10=0, x-a x i s$ and line $x=3, x=4$
$\begin{aligned} &\int_{3}^{4}|y| d x=\int_{3}^{4}\left(\frac{3 x+10}{x-2}\right) d x \\\\ &=\int_{3}^{4}\left(\frac{3 x-6+16}{x-2}\right) d x \\\\ &=\int_{3}^{4}\left(3+\frac{16}{x-2}\right) d x \end{aligned}$
$\begin{aligned} &=[3 x+16 \log |x-2|]_{3}^{4} \\\\ &=12+16 \log |2|-9-16 \log |1| \\\\ &=16 \log 2+3 \text { sq.unit } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 21

Answer:
$\frac{\pi }{2}(\pi +2) s q \cdot { unit }$
Hint:
use definite integral.
Given:
Draw rough sketch of curve $y=\frac{\pi}{2}+2 \sin ^{2} x$ Find area between $x-axis$ the curve and ordinate $x=0,x=\pi$
Solution:

$y=\frac{\pi}{2}+2 \sin ^{2} x$ is an arc cutting $y-axis$ at $\left ( 1.57,0 \right )$

_$x=\pi$at_{$(\pi ,1.57)$}

Area of shaded region
$\begin{aligned} &A=\int_{0}^{\pi}|y| d x \\\\ &A=\int_{0}^{\pi} y d x \\\\ &A=\int_{0}^{\pi}\left[\frac{\pi}{2}+2 \sin ^{2} x\right] d x \end{aligned}$
$\begin{aligned} &A=\int_{0}^{\pi}\left[\frac{\pi}{2}+2\left(\frac{1-\cos 2 x}{2}\right)\right] d x \\\\ &A=\int_{0}^{\pi}\left[\frac{\pi}{2}\right] d x+\int_{0}^{\pi}[(1-\cos 2 x)] d x \\\\ &A=\frac{\pi}{2}[x]_{0}^{\pi}+\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi} \end{aligned}$
$\begin{aligned} &A=\frac{\pi}{2}[\pi]+\left[\pi-\frac{\sin 2 \pi}{2}-0\right] \\\\ &A=\pi\left(\frac{\pi}{2}+1\right) \\\\ &A=\pi\left(\frac{\pi+2}{2}\right) \end{aligned}$
$A=\frac{\pi }{2}(\pi +2) s q \cdot{ unit }$

Areas of Bounded Region exercise 20.1 question 22

Answer:
$\frac{3\pi }{2} s q \text { unit }$
Hint:
You know about integral of $\sin ^{2}x$
Given:
Draw rough curve $y=\frac{x}{\pi}+2 \sin ^{2} x$ . Find area between $x-axis$ with ordinate $x=0, x=\pi$
Solution:

$A=\int_{0}^{\pi}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]$
$\begin{aligned} &A=\int_{0}^{\pi}\left(\frac{x}{\pi}+2 \sin ^{2} x\right) d x \\\\ &A=\frac{1}{\pi} \int_{0}^{\pi}(x) d x+2 \int_{0}^{\pi}\left(\sin ^{2} x\right) d x \\\\ &A=\frac{1}{\pi}\left[\frac{x^{2}}{2}\right]_{0}^{\pi}+2\left[\frac{x}{2}-\frac{1}{2} \sin x \cos x\right] \end{aligned}$
$\begin{aligned} &A=\frac{\pi^{2}}{2 \pi}+\frac{2}{2}\left[\pi-\frac{1}{2} \sin x \cos x-0\right] \\\\ &A=\frac{\pi}{2}+\pi \\\\ &A=\frac{3 \pi}{2} s q \cdot u n i t \end{aligned}$

Areas of Bounded Region exercise 20.1 question 23

Answer:
$4\; s q \cdot { unit }$
Hint:
Curve, $y=\cos x$
Given:
Find area by curve $y=\cos x$ and ordinate $x=0$ and $x=2\pi$
Solution:

$\begin{aligned} &A=\int_{0}^{2 \pi}|y| d x \\\\ &A=\int_{0}^{\frac{\pi}{2}}|y| d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}|y| d x+\int_{\frac{3 \pi}{2}}^{2 \pi}|y| d x \end{aligned}$
$\begin{aligned} &A=\int_{0}^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}-y d x+\int_{\frac{3 \pi}{2}}^{2 \pi} y d x \\\\ &A=\int_{0}^{\frac{\pi}{2}} \cos x d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}-\cos x d x+\int_{\frac{3 \pi}{2}}^{2 \pi} \cos x d x \end{aligned}$
$\begin{aligned} &A=[\sin x]_{0}^{\frac{\pi}{2}}+[-\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}+[\sin x]_{ \frac{3 \pi}{2}}^{2\pi} \\\\ &A=1+[1+1]+0-(-1) \\\\ &A=4 \; \mathrm{sq} \cdot \mathrm{unit} \end{aligned}$

Areas of Bounded Region exercise 20.1 question 24

Answer:
$y=\sin x, y=\sin 2 x \text { is in ratio } 2: 3$
Hint:
Use two graph of $\sin x$.
Given:
Show that area under curve $y=\sin x, y=\sin 2 x$ between $x=0$ and $x=\frac{\pi }{3}$ are in ratio $2:3$
Solution:


Area of graph 1 :
$A_{1}=\int_{0}^{\frac{\pi}{3}}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]$
$\begin{aligned} &A_{1}=\int_{0}^{\frac{\pi}{3}} y d x \\\\ &A_{1}=\int_{0}^{\frac{\pi}{3}} \sin x d x \\\\ &A_{1}=[-\cos x]_{0}^{\frac{\pi}{3}} \end{aligned}$
$\begin{aligned} &A_{1}=\left[-\cos \frac{\pi}{3}+\cos 0\right] \\\\ &A_{1}=-\frac{1}{2}+1 \\\\ &A_{1}=\frac{1}{2} \end{aligned}$
Area of graph 2 : ............(i)
$A_{2}=\int_{0}^{\frac{\pi}{3}}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]$
$\begin{aligned} &A_{2}=\int_{0}^{\frac{\pi}{3}} y d x \\\\ &A_{2}=\int_{0}^{\frac{\pi}{3}} \sin 2 x d x \\\\ &A_{2}=\int_{0}^{\frac{\pi}{3}} 2 \sin x \cos x d x \end{aligned}$
$\begin{aligned} &A_{2}=2\left[\frac{-1}{4} \cos 2 x\right]_{0}^{\frac{\pi}{3}} \\\\ &A_{2}=\frac{1}{2}\left[-\cos 2 \frac{\pi}{3}+\cos 0\right] \end{aligned}$
$\begin{aligned} &A_{2}=\frac{1}{2}\left[1+\frac{1}{2}\right] \\\\ &A_{2}=\frac{1}{2}\left[\frac{3}{2}\right] \\\\ &A_{2}=\frac{3}{4} \end{aligned}$ ............(ii)
From (i) and (ii)
$\frac{A_{1}}{A_{2}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$
Thus area of curve $y=\sin x, y=\sin 2 x$ for $x=0$ and $x=\frac{\pi }{3}$ are in ratio $2:3$

Areas of Bounded Region exercise 20.1 question 25

Answer:
Each equal to$\frac{\pi }{2} s q \text { unit }$
Hint:
Use $\sin ^{2} x, \cos ^{2} x$
Given:
Compare area under curve$y=\cos ^{2} x, y=\sin ^{2} x$ between$x=0, x=\pi$
Solution:
Apply reduction formula,
$\begin{aligned} &\int \cos ^{n} x d x=\frac{n-1}{n} \int \cos ^{n-2} x d x+\frac{\cos ^{n-1} x \sin x}{n} \\\\ &A_{1}=2\left[\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} d x+\frac{\sin x \cos x}{2}\right]_{\frac{\pi}{2}}^{\pi} \end{aligned}$
$\begin{aligned} &A_{1}=[x+\sin x \cos x]_{\frac{\pi}{2}}^{\pi} \\\\ &A_{1}=[\pi+\sin \pi \cos \pi]-\left[\frac{\pi}{2}+\sin \frac{\pi}{2} \cos \frac{\pi}{2}\right] \\\\ &A_{1}=\frac{\pi}{2} \end{aligned}$
Apply reduction formula,
$\int \sin ^{n} x d x=\frac{n-1}{n} \int \sin ^{n-2} x d x+\frac{\sin ^{n-1} x \cos x}{n}$
$\begin{aligned} &A_{2}=\left[\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} d x-\frac{\sin x \cos x}{2}\right]_{0}^{\pi} \\\\ &A_{2}=\frac{1}{2}[x-\sin x \cos x]_{0}^{\pi} \end{aligned}$
$\begin{aligned} &A_{2}=\frac{1}{2}[\pi-\sin \pi \cos \pi]-\frac{1}{2}[0+\sin 0 \cos 0] \\\\ &A_{2}=\frac{\pi}{2} \\\\ &A_{1}=A_{2} \end{aligned}$

Areas of Bounded Region exercise 20.1 question 26

Answer:
$a b\left\{e \sqrt{1-e^{2}}+\sin ^{-1} e\right\}$
Hint:
Use ellipse formula
Given:
Find area bounded by ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and ordinate $x=ae$ and $x=0$ where$b^{2}=a^{2}\left(1-e^{2}\right)$ and $e< 1$
Solution:

Required area$=$ Area of region
$=$ Area of$BORQSP$
$\begin{aligned} &=2 \times \text { Area OBPS } \\\\ &=2 \int_{0}^{a e} y d x \end{aligned}$
We know that,
$\begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\\\ &\frac{y^{2}}{b^{2}}=\frac{a^{2}-x^{2}}{a^{2}} \\\\ &y^{2}=\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right) \end{aligned}$$\begin{aligned} &y=\pm \sqrt{\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right)} \\\\ &y=\pm \frac{b}{a} \sqrt{a^{2}-x^{2}} \end{aligned}$

Since _$OBPS$in first quadrant, value of_$y$ is positive

$\begin{aligned} &y=\frac{b}{a} \sqrt{a^{2}-x^{2}} \\\\ &=2 \int_{0}^{a e} y d x \end{aligned}$
Required area $=$
$\begin{aligned} &=2 \int_{0}^{a e} \frac{b}{a} \sqrt{a^{2}-x^{2}} d x \\\\ \end{aligned}$
$=\frac{2 b}{a} \int_{0}^{a e} \sqrt{a^{2}-x^{2}} d x \quad\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]$
$\begin{aligned} &=\frac{2 b}{a}\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a e} \\\\ &=\frac{2 b}{a}\left[\frac{a e}{2} \sqrt{a^{2}-(a e)^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{a e}{a}\right]-\frac{2 b}{a}\left[\frac{0}{2} \sqrt{a^{2}-0}+\frac{a^{2}}{2} \sin ^{-1} \frac{0}{a}\right] \end{aligned}$
$\begin{aligned} &=\frac{2 b}{a}\left[\frac{a e}{2} a \sqrt{1-e^{2}}+\frac{a^{2}}{2} \sin ^{-1} e-0\right] \\\\ &=\frac{2 b}{a}\left[\frac{a^{2}}{2} e \sqrt{1-e^{2}}+\frac{a^{2}}{2} \sin ^{-1} e\right] \end{aligned}$
$\begin{aligned} &=\frac{2 b}{a} \times \frac{a^{2}}{2}\left[e \sqrt{1-e^{2}}+\sin ^{-1} e\right] \\\\ &=a b\left[e \sqrt{1-e^{2}}+\sin ^{-1} e\right] \end{aligned}$

Areas of Bounded Region exercise 20.1 question 27

Answer:
$\frac{a^{2}}{12}(4 \pi-3 \sqrt{3}) { sq } \cdot { unit }$
$96\; s q \cdot { unit }$
Hint:
Use definite integrals.
Given:
Find area of circle $x^{2}+y^{2}=a^{2}$ cut by the line $x=\frac{a}{2}$
Solution:

By solving equation $\frac{a}{2}, \frac{\sqrt{3} a}{2}$and $\frac{a}{2}, \frac{-\sqrt{3} a}{2}$
Hence form diagram, we get
Required area $=2 \times \text { Area of } \mathrm{AOB}$
$\begin{aligned} &=2 \int_{a / 2}^{a} \sqrt{a^{2}-x^{2}} d x \\\\ &=2\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{a / 2}^{a} \end{aligned}$
$\begin{aligned} &=2\left[\frac{a^{2}}{2}\left(\frac{\pi}{2}\right)-\frac{a}{4}\left(\frac{a \sqrt{3}}{2}\right)-\frac{a^{2}}{2}\left(\frac{\pi}{6}\right)\right] \\\\ &=\frac{a^{2}}{12}[6 \pi-3 \sqrt{3}-2 \pi] \\\\ &=\frac{a^{2}}{12}[4 \pi-3 \sqrt{3}] { sq\;. unit } \end{aligned}$

Areas of Bounded Region exercise 20.1 question 28

Answer:
$\frac{56}{3}a^{2}\; s q\cdot { unit }$
Hint:
Use definite integral.
Given:
Find area of curve $x=a t^{2}, y=2 a t$ between ordinate $t=1$ and $t=2$
Solution:

Given $x=a t^{2}, y=2 a t$
Required area$=2 \times \text { Area of } A B C D$
$\begin{aligned} &=2 \int_{a}^{4 a} y d x \\\\ &=2 \times 2 \int_{a}^{4 a} \sqrt{a x} d x \end{aligned}$
$\begin{aligned} &=8 \sqrt{a}\left[\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{a}^{4 a} \\\\ &=\frac{56}{3} a^{2}\; s q \cdot u n i t \end{aligned}$

Areas of Bounded Region exercise 20.1 question 29

Answer:
$6\pi \; s q \cdot { unit }$
Hint:
Use $x=3 \cos t, y=2 \sin t$
Given:
Find area enclosed by curve _{$x=3 \cos t, y=2 \sin t$}
Solution:

The given curve $x = 3cot\; t, y = 2sin\; t$ represents the parametric equation of ellipse.
Eliminating the parameter $t$, we get,
$\frac{x^{2}}{9}+\frac{y^{2}}{4}=\cos ^{2} t+\sin ^{2} t=1$
This represent the Cartesian equation of the ellipse with centre $\left ( 0,0 \right )$. The co ordinates of the vertices are $\left ( \pm 3,0 \right )$ and $\left ( 0,\pm 2 \right )$.
$\therefore$ Required area $=$ Area of the shaded region
$=$ Area of region$OABO$
$\begin{aligned} &=4 \times \int_{0}^{3} y_{\text {ellipse }} d x \\\\ &=4 \times \int_{0}^{3} \sqrt{4\left(1-\frac{x^{2}}{9}\right)} d x \\\\ &=4 \times \frac{2}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x \end{aligned}$
$\begin{aligned} &=\frac{8}{3}\left(\frac{x}{2} \sqrt{9-x}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right)_{0}^{3} \\\\ &=\frac{8}{3}\left[\left(0+\frac{9}{2} \sin ^{-1} 1\right)-(0+0)\right] \\\\ &=\frac{8}{3} \times \frac{9}{2} \times \frac{\pi}{2} \end{aligned}$
$=6\pi \; s q \cdot { unit }$

Areas of Bounded Region exercise 20.1 question 30

Answer:
$16^{\frac{1}{3}}$
Hint:
Use integration.
Given:
If area between curve $x=y^{2}$ and $x=4$ divide two equal part of line $x=a$ .find using integration.
Solution:

Given curve

_$y^{2}=x$
Let $AB$ represent line $x=a$
$CD$ represent line $x=4$
Since line $x=a$ divide the region in two equal parts
Area of $OBA=$ Area of$ABCD$
$\begin{aligned} &2 \int_{0}^{a} y d x=2 \int_{a}^{4} y d x \\\\ &\int_{0}^{a} y d x=\int_{a}^{4} y d x \end{aligned}$ .............(i)
Now, $y^{2}=x \Rightarrow y=\pm \sqrt{x}$
$y=\sqrt{x}$
From (i)
$\int_{0}^{a} y d x=\int_{a}^{4} y d x$
$\begin{aligned} &\int_{0}^{a} \sqrt{x} d x=\int_{a}^{4} \sqrt{x} d x \\\\ &\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_{0}^{a}=\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_{a}^{4} \end{aligned}$
$\begin{aligned} &{\left[x^{\frac{3}{2}}\right]_{0}^{a}=\left[x^{\frac{3}{2}}\right]_{a}^{4}} \\\\ &a^{\frac{3}{2}}=4^{\frac{3}{2}}-a^{\frac{3}{2}} \\\\ &2 a^{\frac{3}{2}}=4^{\frac{3}{2}} \end{aligned}$
Take $\frac{2}{3}^{t h}$root on both sides
$\begin{aligned} &2^{\frac{2}{3}} \cdot a^{\frac{3}{2} \cdot \frac{2}{3}}=4^{\frac{3}{2} \frac{2}{3}} \\\\ &2^{\frac{2}{3}} \cdot a=4 \\\\ &a=\frac{2^{2}}{2^{\frac{2}{3}}} \end{aligned}$
$\begin{aligned} &a=2^{2-\frac{2}{3}}=2^{\frac{6-2}{3}} \\\\ &a=2^{\frac{4}{3}} \\\\ &a=\left(2^{2}\right)^{\frac{2}{3}}=4^{\frac{2}{3}} \end{aligned}$
$\begin{aligned} &=\left(4^{2}\right)^{\frac{1}{3}}=16^{\frac{1}{3}} \\\\ &a=16^{\frac{1}{3}} \end{aligned}$