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

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

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 03:19 PM IST

RD Sharma books are the benchmark of CBSE Maths in India. These books are renowned for their detailed and concept-rich chapters. They cover all aspects of the chapter and are the best medium of preparation for CBSE students. Once students prepare from this book, they can rest assured that they have covered all concepts in detail.

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.

## 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

$\frac{8}{3}a^{2}\; sq\cdot units$
Hint:
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

$\frac{32}{3}sq\cdot units$
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

$8\sqrt{3}\; sq\cdot units$
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

$\frac{16}{3}sq\cdot units$
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

$\frac{2}{3}\left(5^{\frac{3}{2}}-1\right)$
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

$\frac{56}{9}sq\cdot units$
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

$\frac{4}{3}sq\cdot unit$
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

$3\pi \; sq\cdot unit$
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

$6\pi \; sq\cdot unit$
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

$\frac{\pi}{2} \; sq\cdot unit$
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

$\frac{\pi a^{2}}{4} s q \cdot \text { unit }$
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

$\pi a^{2}\; s q \cdot \text { unit }$
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

$\frac{27}{2} s q \cdot \text { units }$
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

$\frac{9}{2}\; s q \cdot unit$
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

$9\; s q \cdot \text { units }$
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

$9\; s q \cdot { units }$
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

$3+16\log 2\; s q \cdot { unit }$
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

$\frac{\pi }{2}(\pi +2) s q \cdot { unit }$
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 $\left ( 1.57,0 \right )$

$x=\pi$at$(\pi ,1.57)$

\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

$\frac{3\pi }{2} s q \text { unit }$
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

$4\; s q \cdot { unit }$
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

$y=\sin x, y=\sin 2 x \text { is in ratio } 2: 3$
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

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

$a b\left\{e \sqrt{1-e^{2}}+\sin ^{-1} e\right\}$
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

$\frac{a^{2}}{12}(4 \pi-3 \sqrt{3}) { sq } \cdot { unit }$
$96\; s q \cdot { unit }$
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

$\frac{56}{3}a^{2}\; s q\cdot { unit }$
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

$6\pi \; s q \cdot { unit }$
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

$16^{\frac{1}{3}}$
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}

RD Sharma Class 12th Exercise 20.1 material is expert-created and complies with the CBSE syllabus. This means that students can refer to the solutions to finish their homework as well as prepare for exams. Additionally, as this material is updated to the latest version, students can ensure that all of the questions and their solutions are from the newest version of the book.

RD Sharma books have a history of being comprehensive and detailed. Unfortunately, this means that it contains a large number of unsolved exercise questions. To help students deal with this situation, Career360 has provided RD Sharma Class 12th Exercise 20.1 solutions to ensure that no question is left behind.

Moreover, teachers can't cover the entire syllabus through lectures due to online classes, so they give most of the questions as homework. We at Career360 understand this issue and provide solutions to help students cover the whole syllabus without inconvenience.

RD Sharma Class 12th Exercise 20.1 solutions are meant to guide students with detailed answers. In addition, it is readily available on Career360's website, which makes it even more helpful as students can access it right from their homes using a device with a browser and internet connection.

This material provides different ways to solve a question. Students can refer to it and choose the method that suits them best. Maths is a subject in which every step counts; even one slight mistake in a step can lead to the wrong answer. This is why students are required to practice the sums well and look for all types of problems. This material will help students get plenty of questions for better understanding.

As the solutions are conveniently available to students, it gives them even more confidence to solve questions as they have the option to check their progress with this material. Thousands of students have already started preparing RD Sharma Class 12th Exercise 20.1 material as it is convenient and easy to use. Students should start preparing if they don't yet know about this.

## RD Sharma Chapter-wise Solutions

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

1. Which is better for maths, RD Sharma or NCERT?

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.

2. 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.

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.

4. 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.

5. Do the questions from this material appear in exams?

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.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024