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.

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RD Sharma Solutions Class 12 Mathematics Chapter 20 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 20 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 20 FBQ

Updated on 24 Jan 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.

This Story also Contains

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

The RD Sharma class 12 solutions chapter 20 exercise FBQ is a country-wide popular and most demanded solution book by students and teachers. Students can use the RD Sharma class 12th exercise FBQ for self-practice and take the test and evaluate their scores with its help. Teachers also refer to the RD Sharma class 12th exercise FBQ for reference to assign homework to students and also for the preparation of question papers as most of the question the solution is of the same concept as of the NCERT, which makes it more valuable for students to practice from it and helps them in solving homework without investing much time.

The RD Sharma class 12 chapter 20 exercise FBQ consists of some major concepts that are prepared by experts in the field of mathematics that gives broad explanation and theories of the basic concepts so well that the students find it much efficient to opt for it. Moreover, the RD Sharma class 12th exercise FBQ comes with some expert tips and tricks to solve the questions easily and alternately that the school might not teach.

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