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.

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

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

Areas of Bounded Regions Excercise:FBQ

Areas Of Bounded Region exercise Fill in the blanks question 1

Answer:

$\frac{37}{3}$ 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

Answer:

$\frac{297}{6}$ 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

Answer:

$2c \log_{e}2$ 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

Answer:

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

Answer:

$2 \log 2$
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

Answer:

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 $\begin{aligned} =\int_{0}^{4}(a \sqrt{x}+b x) d x \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \\ \end{aligned}$
$\begin{aligned} &8=\int_{0}^{4} a \sqrt{x} d x+\int_{0}^{4} b x d x \\ &8=\left[\frac{a x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4}+\left[\frac{b x^{2}}{2}\right]_{0}^{4} \\ \end{aligned}$
$\begin{aligned} &8=\frac{2}{3} a\left[(4)^{\frac{3}{2}}-0\right]+\frac{b}{2}\left[(4)^{2}-0\right] \\ &8=\frac{2}{3} \times 8 a+\frac{b}{2} \times 16 \\ &8=\frac{16 a}{3}+\frac{16 b}{2} \\ \end{aligned}$
$\begin{aligned} &\frac{1}{2}=\frac{a}{3}+\frac{b}{2} \\ &\frac{6}{2}=2 a+3 b \\ &3=2 a+3 b \end{aligned}$

Areas Of Bounded Region exercise Fill in the blanks question 7
Answer:

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

Answer:

$\frac{64}{3}$ 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

Answer:

$\frac{8a^{2}}{3}$ 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

Answer:

$\frac{1}{\sqrt{3}}$
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

Answer:

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

Answer:

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

Answer:

$\pi$ 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

Answer:

$20\pi$ 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

Answer:

$\frac{7}{2}sq.units$
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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