RD Sharma Class 12 Exercise 20.1 Areas of bounded region Solutions Maths

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RD Sharma books are more detailed and contain exam-oriented questions. NCERT books are useful for basic education, but they don't get to the level of RD Sharma in the case of content.

Q: Can I solve NCERT questions after preparing this material?

As RD Sharma books have a slight edge over NCERT materials, students can definitely solve NCERT questions after preparing from RD Sharma Class 12 Chapter 20 Exercise 20.1 material.

Q: What is a bounded region?

Any curve or figure on a graph that is bounded on all sides is called a bounded region. It can also be an outcome of the intersection of two curves. For more information, check Class 12 RD Sharma Chapter 20 Exercise 20.1 Solutions.

Q: How to calculate the area of a bounded region?

The area of a bounded region can be calculated using the integral of its function after applying the horizontal or vertical limits of the region. To know more, refer to RD Sharma Class 12 Solutions Areas of Bounded Region Ex 20.1.

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As RD Sharma books are widely used in CBSE schools, questions from RD Sharma Class 12 Solutions Chapter 20 Ex 20.1 could appear in exams.

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RD Sharma Class 12 Solutions Chapter20 Areas Of Bounded Region - Other Exercise
Areas of Bounded Regions Excercise: 20.1
RD Sharma Chapter-wise Solutions

RD Sharma Class 12th Exercise 20.1 of Indefinite Integrals contains 30 questions, 22 of which are Level 1 and eight are Level 2. The Level 1 sums are relatively easier and can be completed swiftly. Level 2 sums, however, require some conceptual understanding and are slightly longer. RD Sharma solutions These questions are based on the area of the region bounded between line and parabola, curve and line, and many more by using integration and also draw a rough sketch of the graph of the function then evaluate the region. Students are suggested to complete this exercise efficiently with the help of the material provided by Career360.

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

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

We get

Consider $AB$ as line intersecting the $x-axis$ at point

So required area

Substituting the value of,

Substituting the value of ,

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

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 inand determine the area of region enclosed by curve, theand line
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:
and evaluate What represent of graph
Solution:

Let's draw graph,

Required area

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 curveand line
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:

$x$	$0$	$\frac{\pi }{6}$	$\frac{\pi }{2}$	$\frac{5\pi }{6}$	$\pi$
$sin \; x$	$0$	$\frac{1}{2}$	$1$	$\frac{1}{2}$	$0$
$2\sin ^{2}x$	$1.57$	$2.07$	$3.57$	$2.07$	$1.5$

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

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:

$x$	$0$	$\frac{\pi }{6}$	$\frac{\pi }{2}$	$\frac{5\pi }{6}$	$\pi$
$\sin x$	$0$	$\frac{1}{2}$	$1$	$\frac{1}{2}$	$0$
$\frac{x}{\pi }+2\sin ^{2}x$	$0$	$\frac{2}{3}$	$\frac{5}{2}$	$\frac{4}{3}$	$1$

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

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:

$x$	$0$	$\frac{\pi }{6}$	$\frac{\pi }{4}$	$\frac{\pi }{3}$	$\frac{\pi }{2}$
$\cos ^{2}x$	$1$	$0.5$	$0.75$	$0.25$	$0$
$\sin ^{2}x$	$0$	$0.25$	$0.5$	$0.75$	$1$

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 in first quadrant, value of 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
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

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
$\begin{aligned} &2^{\frac{2}{3}} \cdot a^{\frac{3}{2} \cdot \frac{2}{3}}=4^{\frac{3}{2} \frac{2}{3}} \\\\ &2^{\frac{2}{3}} \cdot a=4 \\\\ &a=\frac{2^{2}}{2^{\frac{2}{3}}} \end{aligned}$
$\begin{aligned} &a=2^{2-\frac{2}{3}}=2^{\frac{6-2}{3}} \\\\ &a=2^{\frac{4}{3}} \\\\ &a=\left(2^{2}\right)^{\frac{2}{3}}=4^{\frac{2}{3}} \end{aligned}$
$\begin{aligned} &=\left(4^{2}\right)^{\frac{1}{3}}=16^{\frac{1}{3}} \\\\ &a=16^{\frac{1}{3}} \end{aligned}$

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3. What is a bounded region?

Any curve or figure on a graph that is bounded on all sides is called a bounded region. It can also be an outcome of the intersection of two curves. For more information, check Class 12 RD Sharma Chapter 20 Exercise 20.1 Solutions.

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