RD Sharma Solutions Class 12 Mathematics Chapter 20 FBQ

# RD Sharma Solutions Class 12 Mathematics Chapter 20 FBQ

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

The Class 12 RD Sharma chapter 20 exercise FBQ solution deals with the chapter of 'Area of bounded region,' which brings out the concepts of a bounded region, which means any flat, curved, or irregular expanse of a surface or the extent of a two-dimensional surface enclosed within a specified boundary or geometric figure. The RD Sharma class 12 exercise FBQ collects the most critical questions important in this chapter.

## Areas of Bounded Regions Excercise:FBQ

Areas Of Bounded Region exercise Fill in the blanks question 1

$\frac{37}{3}$ Sq. units
Hint:
Use indefinite integral formula then put limits to solve this integral.
Given:
$x=y^{2},y-axis$
Explanation:
$y=3,x=9\;\;\;\;\;\;\;\;\;\;\;\; \left [ \because x=y^{2} \right] \\ y=4,x=16$

$y^{2}=x$ denotes the curve in the graph.
Area, $OABC=\int_{3}^{4}y^{2}dy$
\begin{aligned} &=\left[\frac{y^{3}}{3}\right]_{3}^{4}\\ &=\frac{1}{3}\left[(4)^{3}-(3)^{3}\right] & &&&&&&& \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]\\ &=\frac{1}{3}[64-27]\\ &=\frac{37}{3} \mathrm{sq} \cdot \text { units } \end{aligned}

Areas Of Bounded Region exercise Fill in the blanks question 2

$\frac{297}{6}$ sq.units
Hint:
Use indefinite integral formula then put limits to solve this integral
Given:
$y=x^{2}+x,x-axis\\ x=2,x=5$
Explanation:
Required area,
\begin{aligned} &=\int_{2}^{5}\left(x^{2}+x\right) d x \\ &=\left[\frac{x^{3}}{3}+\frac{x^{2}}{2}\right]_{2}^{5} \\ \end{aligned}
\begin{aligned}&=\left[\frac{(5)^{3}}{3}+\frac{(5)^{2}}{2}\right]-\left[\frac{(2)^{3}}{3}+\frac{(2)^{2}}{2}\right]&&&&&&&& \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}\right] \\ \end{aligned}
\begin{aligned} &=\frac{125}{3}-\frac{8}{3}+\frac{25}{2}-\frac{4}{2} \\ &=\frac{117}{3}+\frac{21}{2} \\ \end{aligned}
$=\frac{297}{6}$ sq.units

Areas Of Bounded Region exercise Fill in the blanks question 3

$2c \log_{e}2$ sq. units
Hint:
Use indefinite integral formula then put limits to solve this integral
Given:
$xy=c,x-axis\\x=1,x=4$
Explanation
Required area,
\begin{aligned} &=\int_{1}^{4} \frac{c}{x} d x\\ \end{aligned}
\begin{aligned} &\begin{aligned} &=c\left[\log _{e} x\right]_{1}^{4} \\ &=c\left[\log _{e} 4-\log _{e} 1\right] \\ &=c \log _{e} 4 \end{aligned} \quad\left[\begin{array}{l} \because \int_{a}^{b} \frac{1}{x} d x=[\log x]_{a}^{b} \\ \because \log _{e} 1=0 \end{array}\right]\\ \end{aligned}
\begin{aligned} &=c \log _{e} 2^{2}\\ &=2 c \log _{e} 2 \mathrm{sq} \cdot \text { units } \end{aligned}

Areas Of Bounded Region exercise Fill in the blanks question 4

4 sq.units
Hint:
Use indefinite integral formula then put limits to solve this integral
Given:
$y=\sin x, x-axis\\ x=0, x=2\pi$
Explanation:
Required area=$2\int_{0}^{\pi} \sin x\;\; dx$
$=\left [ -2 \cos x \right ]_{0}^{\pi}\\ =-2\left [ \cos \pi-\cos 0 \right ]\\ =-2\left [ -1-1 \right ]\\ =4 sq. units\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \begin{bmatrix} \because \int_{a}^{b} \sin x\;\;dx=[-cos x ]_{a}^{b}\\ \because \cos \pi=-1 \\ \because \cos 0= 1 \end{bmatrix}$

Areas Of Bounded Region exercise Fill in the blanks question 5

$2 \log 2$
Hint:
Use indefinite integral formula then put limits to solve this integral
Given:
$y=\tan x, x-axis\\ x=\frac{-\pi}{3}, x=\frac{\pi}{3}$
Explanation:
\begin{aligned} &\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \tan x d x=2 \int_{0}^{\frac{\pi}{3}} \tan x d x\\ &=[-2 \log \cos x]_{0}^{\frac{\pi}{3}}\\ \end{aligned}
\begin{aligned} &\begin{aligned} &=-2\left[\log \cos \frac{\pi}{3}-\log \cos 0\right] \\ &=-2\left[\log \frac{1}{2}-0\right] \\ &=-2 \log \frac{1}{2} \\ &=-2[\log 1-\log 2] \end{aligned} \quad\left[\begin{array}{l} \because \int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x \\ \because \int \tan x d x=-\log \cos x+c \\ \because[\log m-\log n]=\log \frac{m}{n} \\ \because \log 1=0 \end{array}\right]\\ &=2 \log 2 s q \cdot \text { units } \end{aligned}

Areas Of Bounded Region exercise Fill in the blanks question 6

3
Hint:
Use indefinite integral formula then put limits to solve this integral
$y=a\sqrt{x}+bx,x-asix\\ x=0, x=4\\$
Area=8 sq. units
Given:
Explanation:
Area \begin{aligned} =\int_{0}^{4}(a \sqrt{x}+b x) d x \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \\ \end{aligned}
\begin{aligned} &8=\int_{0}^{4} a \sqrt{x} d x+\int_{0}^{4} b x d x \\ &8=\left[\frac{a x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4}+\left[\frac{b x^{2}}{2}\right]_{0}^{4} \\ \end{aligned}
\begin{aligned} &8=\frac{2}{3} a\left[(4)^{\frac{3}{2}}-0\right]+\frac{b}{2}\left[(4)^{2}-0\right] \\ &8=\frac{2}{3} \times 8 a+\frac{b}{2} \times 16 \\ &8=\frac{16 a}{3}+\frac{16 b}{2} \\ \end{aligned}
\begin{aligned} &\frac{1}{2}=\frac{a}{3}+\frac{b}{2} \\ &\frac{6}{2}=2 a+3 b \\ &3=2 a+3 b \end{aligned}
1
Hint:
Use indefinite integral formula then put limits to solve this integral
Given:
$y=2^{kx},x=0,x=2\\$
Area= $3 \log_{2}e$
Explanation:Area \begin{aligned} =\int_{0}^{2} 2^{k x} d x \quad\left[\because \int a^{x} d x=\frac{a^{x}}{\log _{e} a}\right] \\ \end{aligned}
\begin{aligned} &3 \log _{2} e=\left[\frac{1}{k} \frac{2^{k x}}{\log _{e} 2}\right]_{0}^{2} \\ \end{aligned}
\begin{aligned} &3 \log _{2} e=\left[\frac{\log _{2} e}{k} 2^{k x}\right]_{0}^{2} \quad\left[\because \log _{a} e=\frac{1}{\log _{e} a}\right] \\ \end{aligned}
\begin{aligned} &3 k \log _{2} e=\log _{2} e\left[2^{k x}\right]_{0}^{2} \\ &3 k=\left(2^{k}\right)^{2}-\left(2^{k}\right)^{0} \\ &3 k=2^{2 k}-1 \\ &2^{2 k}-3 k=1 \end{aligned}

Areas Of Bounded Region exercise Fill in the blanks question 8

$\frac{64}{3}$sq. units
Hint:
Use indefinite integral formula then put limits to solve this integral
Given:
$y^{2}=x,y=4,y-axis(x=0)$
Explanation:
$x=0\\ \Rightarrow y^{2}=x\Rightarrow y=0$
Required area
\begin{aligned} &=\int_{0}^{4} y^{2} d y \\ &=\left[\frac{y^{3}}{3}\right]_{0}^{4}\;\;\;\;\;\;\;\;\;\;\;\;\; \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \\ &=\frac{(4)^{3}-(0)^{3}}{3}\\ \end{aligned}
$=\frac{64}{3}$ sq.units

Areas Of Bounded Region exercise Fill in the blanks question 9

$\frac{8a^{2}}{3}$ sq.units
Hint:
Use this formula to integrate : $\int_{a}^{b}x^{n}dx=\left [ \frac{x^{n+1}}{n+1} \right ]_{a}^{b}$
Given:$y^{2}=4ac$, latus-rectum
Explanation:
Latus-rectum(x=a)

Area \begin{aligned} =2 \int_{0}^{a} 2 \sqrt{a x} d x \\ \end{aligned}
\begin{aligned}&=2 \times 2\left[\sqrt{a} \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{a} \\ \end{aligned}
\begin{aligned}&=2 \times \frac{4}{3} \sqrt{a}\left[a^{\frac{3}{2}}-0\right] \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \\ \end{aligned}
$=\frac{8a^{2}}{3}$ sq.units

Areas Of Bounded Region exercise Fill in the blanks question 10

$\frac{1}{\sqrt{3}}$
Hint:
Use this formula to integrate : $\int_{b}^{a}x^{n} dx=\left [ \frac{x^{n+1}}{n+1} \right ]_{a}^{b}$
Given:
$y=ax^{2},x=ay^{2},a>0$
Area=1sq. units
Explanation:
Intersection point
$y=ax^{2}=a(ay^{2})^{2}\;\;\;\;\;\;\;\;\;\;\;\;[\because x=ay^{2}]\\ y=a^{3}y^{4}\\ a^{3}y^{4}-y=0\\ y(a^{3}y^{3}-1)=0\\ y=0\;\;\;\;\;\;\;y=\frac{1}{a}\\ x=a\left [ \frac{1}{a} \right ]^{2}=\frac{1}{a}$

Required area= $\int_{1}^{\frac{1}{a}}\left ( \sqrt{\frac{x}{a}} -ax^{2}\right )dx$
\begin{aligned} &=\frac{1}{\sqrt{a}} \int_{0}^{\frac{1}{a}} \sqrt{x} d x-a \int_{0}^{\frac{1}{a}} x^{2} d x \\ \end{aligned}
\begin{aligned} &=\frac{1}{\sqrt{a}}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{\frac{1}{a}}-a\left[\frac{x^{3}}{3}\right]_{0}^{\frac{1}{a}} \\ \end{aligned}
\begin{aligned} &=\frac{2}{3 \sqrt{a}}\left[\left(\frac{1}{a}\right)^{\frac{3}{2}}-0\right]-\frac{a}{3}\left[\left(\frac{1}{a}\right)^{3}-0\right] \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \\ \end{aligned}
\begin{aligned} &=\frac{2}{3 \sqrt{a}}\left(\frac{1}{a \sqrt{a}}\right)-\frac{1}{3 a^{2}} \\ &=\frac{1}{3 a^{2}} \\ \end{aligned}
Given, Area=1
\begin{aligned} &\frac{1}{3 a^{2}}=1 \\ &a^{2}=\frac{1}{3} \\ &a=\pm \frac{1}{\sqrt{3}} \\ &\text { As } a>0, \\ \therefore a=\frac{1}{\sqrt{3}} \end{aligned}

Areas Of Bounded Region exercise Fill in the blanks question 11

1 sq.units
Hint:
Use this formula to integrate : $\int\sin x \;\; dx =-\cos x+c$
Given:
$y=\sin x,x=0,x=\frac{\pi}{2},x-axis$
Explanation:
Area $=\int_{0}^{\frac{\pi}{2}}\sin x \;\;dx$
$=[-\cos x]_{0}^{\frac{\pi}{2}}\\ =-\cos \frac{\pi}{2}+\cos 0\\$
$=1 sq. units\;\;\;\;\;\;\; \left [\because \int \sin x\;\; dx=-\cos x+c \right ]$

Areas Of Bounded Region exercise Fill in the blanks question 12

2 sq.units
Hint:
Use indefinite integral formula then put limits to solve this integral.
Given:
$y=\cos x, x=0, x=\pi$
Explanation:
Required area,

$=2\int_{0}^{\frac{\pi}{2}}\cos x \;\;\;dx\\ =\left [ 2 \sin x \right ]_{0}^{\frac{\pi}{2}}\\ =2\left [ \sin \frac{\pi}{2}-\sin 0 \right ]\\ =2 sq.units\;\;\;\;\;\;\;\;\;\;\;\;\;\left [ \because \int \cos x \;\; dx=\sin x+c \right ]$

Areas Of Bounded Region exercise Fill in the blanks question 13

$\pi$ sq. units
Hint:
Use this formula to integrate : $\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}$
Given:
$x^{2}+y^{2}=1$
Explanation:

Area \begin{aligned} &\text { Area } A B C D=4(\text { Area } O A B) \\ \end{aligned}
\begin{aligned} &=4 \int_{0}^{1} \sqrt{1-x^{2}} d x \\ &=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} \frac{x}{1}\right]_{0}^{1} \\ \end{aligned}\begin{aligned} &=4\left[\frac{1}{2} \sqrt{1-1}+\frac{1}{2} \sin ^{-1} 1-\frac{0}{2} \sqrt{1-0}-\frac{1}{2} \sin ^{-1} 0\right] \quad\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{1}{2} \sin ^{-1} \frac{x}{a}\right] \\ \end{aligned}
\begin{aligned} &=4\left[\frac{1}{2} \times \frac{\pi}{2}\right] \\ &=4 \times \frac{\pi}{4} \\ &=\pi \text { sq.units } \end{aligned}

Areas Of Bounded Region exercise Fill in the blanks question 14

$20\pi$sq.units
Hint:
Use this formula to integrate: $\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}$
Given:
$x^{2}+y^{2}=1$
Explanation:

Area ABCD=4(Area OAB) … (i)
Now, $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$
$\frac{y^{2}}{16}=1-\frac{x^{2}}{25}\\ \frac{y^{2}}{16}=\frac{25-x^{2}}{25}\\ y=\frac{4}{5}\sqrt{25-x^{2}}$
Now, from (i) we have
Required area
$=4\int_{5}^{0}\frac{4}{5}\sqrt{25-x^{2}}dx\\ =\frac{16}{5}\left [ \frac{x}{2}\sqrt{25-x^{2}}+\frac{25}{2}\sin^{-1}\frac{x}{5} \right ]_{0}^{5}\\ =\frac{16}{5}\left [ \frac{5}{2}\sqrt{25-5^{2}}+\frac{25}{2}\sin^{-1}1 \right ]-\frac{16}{5}\left [ \frac{0}{2}\sqrt{25-0}+\frac{25}{2}\sin^{-1}0 \right ]\\ =\frac{16}{5}\left [ \frac{25}{2}\times \frac{\pi}{2} \right ]\\ =20\pi \;\;sq.units$

Areas Of Bounded Region exercise Fill in the blanks question 15

$\frac{7}{2}sq.units$
Hint:
Use this formula to integrate : $\int_{a}^{b}x^{n}dx=\left [ \frac{x^{n+1}}{n+1} \right ]_{a}^{b}$
Given:
$y=x+1,x-axis,x=2,x=3$
Explanation:
Required area
\begin{aligned} &=\int_{2}^{3}(x+1) d x \\ &=\left[\frac{x^{2}}{2}+x\right]_{2}^{3} \\ \end{aligned}
\begin{aligned} &=\frac{9}{2}+3-\frac{4}{2}-2 &&&&&&&&&&& \quad\left[\begin{array}{l} \left.\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \\ \end{array}\right.\end{aligned}\begin{aligned} &=\frac{5}{2}+1 \\ &=\frac{7}{2} s q \cdot u n i t s \end{aligned}

The RD Sharma class 12th exercise FBQ consists of a total of 15 questions that are short and precise, covering up the essential concepts of this chapter that are important for board exams. The concepts covered in the RD Sharma class 12 solution of Area of bounded region exercise FBQ are mentioned below-

• Method to find the area between two curves

• The area between two curves using vertical and horizontal stripes

• Area of the region bounded by ellipse

• Area of the region bounded by the curve and the line

• Area of the region bounded by a parabola and latus ractum

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## RD Sharma Chapter-wise Solutions

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1. How many exercises do chapter 20 consist of?

There are only five exercises in this chapter that cover all the essential concepts, thus making it worth self-practice.

2. Is it helpful for the preparation for the board exam?

Yes, it is beneficial for the preparation of board exams as it contains questions similar to the questions asked in the exam.

You can download the free PDF for RD Sharma solutions from the official website of Career360 free of cost.

4. How much is the cost of the RD Sharma class12 chapter 20 solution?

The online PDFs are free of cost and can be downloaded through any device from the website of Career360.

5. Can I take the help of the RD Sharma solution to solve homework?

Yes, as teachers refer to these solutions for assigning homework, it is helpful and less time-consuming.

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