RD Sharma Solutions Class 12 Mathematics Chapter 19 FBQ

# RD Sharma Solutions Class 12 Mathematics Chapter 19 FBQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:20 PM IST

The Class 12 RD Sharma Chapter 19 exercise FBQ solution is one of the most considered textbook solutions when it comes for the preparation of board exams or any public examinations. The RD Sharma class 12th exercise FBQ is used by CBSE students, as they face a lot of challenges while solving questions from NCERT, therefore usage of RD Sharma solutions can help in a lot of ways to make seem easier.

## Definite Integrals Excercise:FBQ

Definite Integrals exercise Very short answer type question 1

Answer: $\frac{\pi }{4}$
Hint: You must know the integration rule of trigonometric functions with its limits.
Given: $\int_{0}^{\frac{\pi}{2}} \sin ^{2} x\; d x$
Solution:
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x \\\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 . d x-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x \; d x \end{aligned}
\begin{aligned} &=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{4}\left[\sin \left(2 \frac{\pi}{2}\right)-\sin 0\right] \end{aligned}
\begin{aligned} &=\frac{\pi}{4}-\frac{1}{4}[0] \\\\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Very short answer type question 2

Answer: $\frac{\pi }{4}$
Hint: You must know the integration rule of trigonometric functions.
Given: $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \; d x$
Solution: $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \; d x$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x \\\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 . d x+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x \; d x \end{aligned}
\begin{aligned} &=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right]+\frac{1}{4}\left[\sin \frac{2 \pi}{2}-\sin 0\right] \\\\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Very short answer type question 3

Answer: $\frac{\pi }{2}$
Hint: You must know the integration rules of trigonometric function with its limits
Given: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \; d x$
Solution: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \; d x$
\begin{aligned} &=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x \\\\ &=\frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 1 \cdot d x-\frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos 2 x\; d x \end{aligned}
$=\frac{1}{2}[x]_{\frac{-\pi}{2}}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{\frac{-\pi}{2}}^{\frac{\pi}{2}}$
\begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right]-\frac{1}{4}\left[\sin \frac{2(\pi)}{2}-\sin \frac{2(-\pi)}{2}\right] \\\\ &=\frac{1}{2}\left[\frac{2 \pi}{2}\right]-\frac{1}{4}[0+0] \\\\ &=\frac{\pi}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 4

Answer: $\frac{\pi }{2}$
Hint: You must know the integration rules of trigonometric function with its limits
Given: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x \; d x$
Solution: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x \; d x$
\begin{aligned} &=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x \\\\ &=\frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 1 . d x+\frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos 2 x\; d x \end{aligned}
\begin{aligned} &=\frac{1}{2}[x]_{\frac{-\pi}{2}}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{\frac{\pi}{2}}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right]+\frac{1}{4}\left[\sin \frac{2(\pi)}{2}+\sin \frac{2(-\pi)}{2}\right] \end{aligned}
\begin{aligned} &=\frac{1}{2}\left[\frac{2 \pi}{2}\right]+\frac{1}{4}[0+0] \\\\ &=\frac{\pi}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 5

Hint: You must know the integration rules of trigonometric function with its limits

Given:$\int_{-\pi}^{\frac{\pi}{2}} \sin ^{3} x\; d x$
Solution: $\int_{-\pi}^{\frac{\pi}{2}} \sin ^{3} x\; d x$
\begin{aligned} &=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \sin x \; d x \\\\ &=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(1-\cos ^{2} x\right) \sin x\; d x \end{aligned}
$=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x\; d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x \sin x\; d x$
We know that $\int \sin x \; d x=-\cos x \text { and } \int \cos x\; d x=\sin x$
\begin{aligned} &=[-\cos x]_{\frac{-\pi}{2}}^{\frac{\pi}{2}}+\left[\frac{-\cos ^{3} x}{3}\right]_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \\\\ &=-\cos \frac{\pi}{2}-\left(-\cos \frac{-\pi}{2}\right)+\left[\frac{\cos ^{3}\left(\frac{\pi}{2}\right)}{3}-\frac{\cos ^{3}\left(\frac{-\pi}{2}\right)}{3}\right] \\\\ &=0+0+0-0 \\\\ &=0 \end{aligned}

Definite Integrals exercise Very short answer type question 6

Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x \cos ^{2} x \; d x$
Solution: $I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x \cos ^{2} x \; d x$
\begin{aligned} &f(x)=x \cos ^{2} x \\\\ &f(-x)=(-x) \cos ^{2}(-x) \\\\ &=-x \cos ^{2} x \\\\ &=-f(x) \end{aligned}
Hence, f(x) is an odd function.
Since,$\int_{-a}^{a} f(x) d x=0$ if f(x) is an odd
$\therefore \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x \cos ^{2} x\; d x=0$

Definite Integrals exercise Very short answer type question 7

Answer: $1-\frac{\pi}{4}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \; d x$
Solution: $\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \; d x$
\begin{aligned} &=\int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) d x \\\\ &=\int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x-\int_{0}^{\frac{\pi}{4}} 1 d x \end{aligned}
\begin{aligned} &=[\tan x]_{0}^{\frac{\pi}{4}}-[x]_{0}^{\frac{\pi}{4}} \\\\ &=\tan \frac{\pi}{4}-\tan 0-\frac{\pi}{4}+0 \end{aligned}
\begin{aligned} &=1-0-\frac{\pi}{4}+0 \\\\ &=1-\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Very short answer type question 8

Answer: $\frac{\pi }{4}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{1} \frac{1}{x^{2}+1} d x$
Solution: $\int_{0}^{1} \frac{1}{x^{2}+1} d x$
\begin{aligned} &=\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\\\ &=\frac{\pi}{4}-0 \\\\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Very short answer type question 9

Answer: $-1$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{-2}^{1} \frac{|x|}{x} d x$
Solution: $\int_{-2}^{1} \frac{|x|}{x} d x$
\begin{aligned} &=\int_{-2}^{0} \frac{-x}{x} d x+\int_{0}^{1} \frac{x}{x} d x \\\\ &=\int_{-2}^{0}-1 d x+\int_{0}^{1} 1 d x \end{aligned}
\begin{aligned} &=-[x]_{-2}^{0}+[x]_{0}^{1} \\\\ &=-[0+2]+[1+0] \\\\ &=-2+1 \\\\ &=-1 \end{aligned}
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\infty} e^{-x} d x$
Solution: $\left[\frac{e^{-x}}{-1}\right]_{0}^{\infty}$
\begin{aligned} &=-\left[\frac{1}{e^{\infty}}-\frac{1}{e^{0}}\right] \\\\ &=0+1 \\\\ &=1 \end{aligned}

Definite Integrals exercise Very short answer type question 11

Answer: $\frac{\pi }{2}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{4} \frac{1}{\sqrt{16-x^{2}}} d x$
Solution: $\int_{0}^{4} \frac{1}{\sqrt{16-x^{2}}} d x$
$=\int_{0}^{4} \frac{1}{\sqrt{(4)^{2}-x^{2}}} d x$
\begin{aligned} &\text { Put } x=4 \sin \theta \quad \theta=\sin ^{-1} \frac{x}{4} \\ &\mathrm{~d} \mathrm{x}=4 \cos \theta\; d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{4} \frac{4 \cos \theta}{\sqrt{16-16 \sin ^{2} \theta}} d \theta \\\\ &=\int_{0}^{4} \frac{4 \cos \theta}{\sqrt{16\left(1-\sin ^{2} \theta\right)}} d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{4} \frac{4 \cos \theta}{4 \sqrt{\cos ^{2} \theta}} d \theta \\\\ &=\int_{0}^{4} \frac{4 \cos \theta}{4 \cos \theta} d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{4} 1 d \theta \\\\ &=[\theta]_{0}^{4}=\left[\operatorname{Sin}^{-1} \frac{x}{4}\right]_{0}^{4} \end{aligned}
\begin{aligned} &=\left[\sin ^{-1} \frac{4}{4}-\sin ^{-1} \frac{0}{4}\right] \\\\ &=\sin ^{-1} 1 \\\\ &=\frac{\pi}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 12

Answer: $\frac{\pi }{12}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{3} \frac{1}{x^{2}+9} d x$
Solution: $\int_{0}^{3} \frac{1}{x^{2}+9} d x$
Use the formula $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$
\begin{aligned} &=\frac{1}{3}\left[\tan ^{-1} \frac{x}{3}\right]_{0}^{3} \\\\ &=\frac{1}{3}\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \end{aligned}
\begin{aligned} &=\frac{1}{3}\left[\frac{\pi}{4}\right] \\\\ &=\frac{\pi}{12} \end{aligned}

Definite Integrals exercise Very short answer type question 13

Answer: $\sqrt{2}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{\pi}{2}} \sqrt{1-\cos 2 x} \; d x$
Solution: $\int_{0}^{\frac{\pi}{2}} \sqrt{1-\cos 2 x} \; d x$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{2 \sin ^{2} x} \; d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{2} \sin x\; d x \end{aligned}
\begin{aligned} &=\sqrt{2}[-\cos x]_{0}^{\frac{\pi}{2}} \\\\ &=\sqrt{2}\left[-\cos \frac{\pi}{2}+\cos 0\right] \\\\ &=\sqrt{2} \end{aligned}

Definite Integrals exercise Very short answer type question 14

Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{\pi}{2}} \log \tan x \; d x$
Solution: $\int_{0}^{\frac{\pi}{2}} \log \tan x \; d x$ .................(i)
$\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \tan \left(\frac{\pi}{2}-x\right) d x$
$=\int_{0}^{\frac{\pi}{2}} \log \cot x \; d x$ ..................(ii)
$2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x+\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x$
$=\int_{0}^{\frac{\pi}{2}}(\log \tan x+\log \cot x) d x$
$=\int_{0}^{\frac{\pi}{2}}(\log \tan x \cdot \cot x) d x \quad[\because \log m+\log n=\log m n]$
$=\int_{0}^{\frac{\pi}{2}}(\log 1) d x \quad[\because \tan x \cdot \cot x=1]$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}(0) d x \quad[\because \log 1=0] \\\\ &=0 \end{aligned}

Definite Integrals exercise Very short answer type question 15

Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{\pi}{2}} \log \left(\frac{3+5 \cos x}{3+5 \sin x}\right) d x$ .............(i)
Solution: $\int_{0}^{\frac{\pi}{2}} \log \left(\frac{3+5 \cos x}{3+5 \sin x}\right) d x$
Property: $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
$\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \frac{3+5 \cos \left(\frac{\pi}{2}-x\right)}{3+5 \sin \left(\frac{\pi}{2}-x\right)} d x$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \log \frac{3+5 \sin x}{3+5 \cos x} d x \\\\ &=-\int_{0}^{\frac{\pi}{2}} \log \frac{3+5 \cos x}{3+5 \sin x} d x \end{aligned} ................(ii)
\begin{aligned} &21=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{3+5 \cos x}{3+5 \sin x}\right) d x+\left[-\int_{0}^{\frac{\pi}{2}} \log \frac{3+5 \cos x}{3+5 \sin x} d x\right] \\\\ &21=0 \\\\ &I=0 \end{aligned}

Definite Integrals exercise Very short answer type question 16

Answer: $\frac{\pi }{4}$
Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x$
Solution: $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x$ ....................(i)
Property: $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
$=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n}\left(\frac{\pi}{2}-x\right)}{\sin ^{n}\left(\frac{\pi}{2}-x\right)+\cos ^{n}\left(\frac{\pi}{2}-x\right)} d x$
$=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x$ ..................(ii)
\begin{aligned} &2 \mathrm{l}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n} x+\cos ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x \\\\ &2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x \end{aligned}
\begin{aligned} &21=[x]_{0}^{\frac{\pi}{2}} \\\\ &2 I=\left[\frac{\pi}{2}-0\right] \\\\ &I=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Very short answer type question 17

Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{0}^{\pi} \cos ^{5} x \; d x$
Solution: Using property $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$
$\mathrm{I}=\int_{0}^{\pi} \cos ^{5} x \; d x=\int_{0}^{\pi} \cos x \cos ^{4} x \; d x$
$=\int_{0}^{\pi} \cos x\left(1-\sin ^{2} x\right)^{2} d x$
\begin{aligned} &\mathrm{Put} \sin \mathrm{x}=\mathrm{t} \\\\ &\cos x \; d x=d t \end{aligned}
\begin{aligned} &\mathrm{I}=\int_{0}^{\pi}\left(1-t^{2}\right)^{2} d x \\\\ &=\int_{0}^{\pi}\left(1+t^{4}-2 t^{2}\right) d x \end{aligned}
$=\left[t+\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]_{0}^{\pi}$
\begin{aligned} &=\left[\sin x+\frac{\sin ^{5} x}{5}-\frac{2 \sin ^{3} x}{3}\right]_{0}^{\pi} \quad\quad\quad\quad[\sin \pi=0, \sin 0=0] \\\\ &I=0 \end{aligned}

Definite Integrals exercise Very short answer type question 18

Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta$

Solution: $I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta$ .............(i)
Let $f(x)=\log \left(\frac{a-\sin \theta}{a+\sin \theta}\right)$
$f(-x)=\log \left(\frac{a-\sin (-\theta)}{a+\sin (-\theta)}\right)$
$=\log \left(\frac{a+\sin \theta}{a-\sin \theta}\right) \quad[\because \sin (-x)=-\sin x]$
\begin{aligned} &=-\log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) \\ &=-\mathrm{f}(\mathrm{x}) \end{aligned}
Hence, f(x) is an odd function.

Since, $\int_{-a}^{a} f(x) d x=0$ if f(x) is an odd.

$\therefore \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta=0$

Definite Integrals exercise Very short answer type question 19

Hint: you must know the rule of integration for function of x

Given: $\int_{-1}^{1} x|x| d x$
Solution: $I=\int_{-1}^{1} x|x| d x$
\begin{aligned} &=\int_{-1}^{0} x(-x) d x+\int_{0}^{1} x(x) d x \\\\ &=\int_{-1}^{0}-x^{2} d x+\int_{0}^{1} x^{2} d x \end{aligned}
\begin{aligned} &=-\frac{1}{3}\left[x^{3}\right]_{-1}^{0}+\frac{1}{3}\left[x^{3}\right]_{0}^{1} \\\\ &=-\frac{1}{3}[0-(-1)]+\frac{1}{3}[1-0] \end{aligned}
\begin{aligned} &=-\frac{1}{3}+\frac{1}{3} \\\\ &=0 \end{aligned}

### Definite Integrals exercise Very short answer type question 20

Answer: $\frac{b-a}{2}$
Hint: you must know the rule of integration for function of x

Given: $\int_{a}^{b} \frac{f(x)}{f(x)+f(a+b-x)} d x$
Solution: $\mathrm{I}=\int_{a}^{b} \frac{f(x)}{f(x)+f(a+b-x)} d x$ ...............(i)
Using property $\int_{a}^{b} f(x)=\int_{a}^{b} f(a+b-x) d x$

$\mathrm{I}=\int_{a}^{b} \frac{f(a+b-x)}{f(a+b-x)+f(a+b-(a+b-x))} d x$
$\mathrm{I}=\int_{a}^{b} \frac{f(a+b-x)}{f(a+b-x)+f(x)} d x$ ................(ii)
\begin{aligned} &2 I=\int_{a}^{b} \frac{f(x)+f(a+b-x)}{f(x)+f(a+b-x)} d x \\\\ &2 I=\int_{a}^{b} 1 . d x \end{aligned}
\begin{aligned} &2 \mathrm{l}=[x]_{a}^{b} \\\\ &2 \mathrm{l}=\mathrm{b}-\mathrm{a} \\\\ &\mathrm{I}=\frac{b-a}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 21

Answer: $\frac{\pi }{4}$
Hint: you must know the rule of integration for function of x
Given:$\int_{0}^{1} \frac{1}{1+x^{2}} d x$
Solution: $\int_{0}^{1} \frac{1}{1+x^{2}} d x$
\begin{aligned} &=\left[\tan ^{-1} x\right]_{0}^{1} \\\\ &={\left[\tan ^{-1} 1-\tan ^{-1} 0\right]} \\\\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Very short answer type question 22

Answer: $\frac{1}{2} \log 2$
Hint: you must know the rule of integration for function of x
Given: $\int_{0}^{\frac{\pi}{4}} \tan x\; d x$
Solution: $\int_{0}^{\frac{\pi}{4}} \tan x\; d x$
\begin{aligned} &=\log [\sec x]_{0}^{\frac{\pi}{4}} \\\\ &=\log \left(\sec \frac{\pi}{4}\right)-\log (\sec 0) \\\\ &=\log |\sqrt{2}|-\log |1| \end{aligned}
\begin{aligned} &=\log \sqrt{2}-0 \\\\ &=\log \sqrt{2} \\\\ &=\log (2)^{\frac{1}{2}} \\\\ &=\frac{1}{2} \log 2 \end{aligned}

Definite Integrals exercise Very short answer type question 23

Answer: $\log \frac{3}{2}$
Hint: you must know the rule of integration for function of x
Given:$\int_{2}^{3} \frac{1}{x} d x$
Solution: $\int_{2}^{3} \frac{1}{x} d x$
\begin{aligned} &=[\log |x|]_{2}^{3} \\\\ &=\log 3-\log 2 \\\\ &=\log \frac{3}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 24

Answer: $\pi$
Hint: you must know the rule of integration
Given:$\int_{0}^{2} \sqrt{4-x^{2}}$
Solution: $\int_{0}^{2} \sqrt{4-x^{2}}$
\begin{aligned} &\text { Put } x=2 \sin \theta \\\\ &d x=2 \cos \theta d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{4-4 \sin ^{2} \theta} \cdot 2 \cos \theta d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 4 \cos ^{2} \theta d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 4 \cdot \frac{1}{2}(1+\cos 2 \theta) d \theta \\\\ &=2\left[\theta+\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}} \end{aligned}
\begin{aligned} &=2\left[\frac{\pi}{2}+\frac{\sin \pi}{2}-0-0\right] \\\\ &=\pi \end{aligned}

Definite Integrals exercise Very short answer type question 25

Answer: $\log _{e} 2$
Hint: you must know the rule of integration for function of x
Given: $\int_{0}^{1} \frac{2 x}{1+x^{2}} \mathrm{dx}$
Solution: $\int_{0}^{1} \frac{2 x}{1+x^{2}} \mathrm{dx}$
Put $1+x^{2}=t$
$2 x \; d x=d t$
\begin{aligned} &=\int_{0}^{1} \frac{d t}{t} \\\\ &=[\log t]_{0}^{1} \end{aligned}
\begin{aligned} &=\left[\log \left(1+x^{2}\right)\right]_{0}^{1} \\\\ &=[\log (1+1)-\log (1+0)] \\\\ &=\log _{e} 2 \end{aligned}

Definite Integrals exercise Very short answer type question 26

Answer: $\frac{1}{2}(e-1)$
Hint: you must know the rule of integration for function of x
Given: $\int_{0}^{1} x e^{x^{2}} d x$
Solution:
Put
\begin{aligned} &x^{2}=t \\\\ &2 x d x=d t \\\\ &x d x=\frac{d t}{2} \end{aligned}
\begin{aligned} &=\frac{1}{2} \int_{0}^{1} e^{t} d t \\\\ &=\frac{1}{2}\left[e^{t}\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\frac{1}{2}\left[e^{x^{2}}\right]_{0}^{1} \\\\ &=\frac{1}{2}\left[e^{1}-e^{0}\right] \\\\ &=\frac{1}{2}[e-1] \end{aligned}

Definite Integrals exercise Very short answer type question 27

Answer: $\frac{1}{2}$
Hint: you must know the rule of integration
Given: $\int_{0}^{\frac{\pi}{4}} \sin 2 x \; d x$
Solution: $\int_{0}^{\frac{\pi}{4}} \sin 2 x \; d x$
\begin{aligned} &=\frac{1}{2}(-\cos 2 x)_{0}^{\frac{\pi}{4}} \\\\ &=-\frac{1}{2}(\cos 2 x)_{0}^{\frac{\pi}{4}} \end{aligned}
\begin{aligned} &=\frac{-1}{2}\left[\cos 2\left(\frac{\pi}{4}\right)-\cos 2(0)\right] \\\\ &=\frac{-1}{2}\left[\cos \frac{\pi}{2}-\cos 0\right] \end{aligned}
\begin{aligned} &=\frac{-1}{2}[0-1] \\\\ &=\frac{1}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 28

Answer: $\log _{e} 2$
Hint: you must know the rule of integration
Given: $\int_{e}^{e^{2}} \frac{1}{x \log x} d x$
Solution: Put $\log x=t$
$\frac{d x}{x}=\mathrm{dt}$
\begin{aligned} &=\int_{e}^{e^{2}} \frac{1}{t} d t \\\\ &=[\log \mid t]_{-e}^{e^{2}} \end{aligned}
\begin{aligned} &=[\log [\log x]]_{e}^{e^{2}} \\\\ &=\log \left(\log e^{2}\right)-\log (\log e) \\\\ &=\log (2 \log e)-\log (1) \\\\ &=\log 2 \end{aligned}

Definite Integrals exercise Very short answer type question 29

Hint: you must know the rule of integration
Given: $\int_{0}^{\frac{\pi}{2}} e^{x}(\sin x-\cos x) d x$
Solution: $\int_{0}^{\frac{\pi}{2}} e^{x}(\sin x-\cos x) d x$
$=-\int_{0}^{\frac{\pi}{2}} e^{x}(\cos x-\sin x) d x$
\begin{aligned} &f(x)=\cos x \\\\ &f^{\prime}(x)=-\sin x \\\\ &\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+c \end{aligned}
\begin{aligned} &\text { we get, } I=-\left[e^{x} \cos x\right]_{0}^{\frac{\pi }{2}} \\\\ &=-e^{\frac{\pi}{2}} \cos \frac{\pi}{2}+e^{0} \cos (0) \\\\ &=0+(1)(1) \\\\ &=1 \end{aligned}

Definite Integrals exercise Very short answer type question 30

Answer: $\frac{1}{2} \log \left(\frac{17}{5}\right)$
Hint: you must know the rule of integration
Given: $\int_{2}^{4} \frac{x}{x^{2}+1} d x$
Solution: Put $x^{2}+1=t$
\begin{aligned} &2 x \; d x=d t \\\\ &x \; d x=\frac{d t}{2} \end{aligned}
\begin{aligned} &\mathrm{I}=\frac{1}{2} \int_{2}^{4} \frac{d t}{t} \\\\ &=\frac{1}{2}[\log |t|]_{2}^{4} \end{aligned}
\begin{aligned} &=\frac{1}{2}\left[\log \left|x^{2}+1\right|\right]_{2}^{4} \\\\ &=\frac{1}{2}[\log (16+1)+\log (4+1)] \\\\ &=\frac{1}{2} \log \left(\frac{17}{5}\right) \end{aligned}

Definite Integrals exercise Very short answer type question 31

Hint: you must know the rule of integration
Given: $\text { If } \int_{0}^{1}\left(3 x^{2}+2 x+k\right) d x=0 \text { find } k$
Solution: $\int_{0}^{1}\left(3 x^{2}+2 x+k\right) d x=0$
\begin{aligned} &{\left[\frac{3 x^{3}}{3}+\frac{2 x^{2}}{2}+k x\right]_{0}^{1}=0} \\\\ &{\left[x^{3}+x^{2}+k x\right]_{0}^{1}=0} \end{aligned}
\begin{aligned} &{[1+1+k-0]=0} \\\\ &k=-2 \end{aligned}

Definite Integrals exercise Very short answer type question 32

Answer: $a=2$
Hint: you must know the rule of integration
Given: $\text { If } \int_{0}^{a} 3 x^{2} d x=8, \text { find } a$
Solution: $\int_{0}^{a} 3 x^{2} d x=8$
\begin{aligned} &{\left[\frac{3 x^{3}}{3}\right]_{0}^{a}=8} \\\\ &{\left[x^{3}\right]_{0}^{a}=8} \end{aligned}
\begin{aligned} &a^{3}-0=8 \\\\ &a=\sqrt[3]{8} \\\\ &a=2 \end{aligned}

Definite Integrals exercise Very short answer type question 33

Answer: $f^{\prime}(x)=x \sin x$
Hint: you must know the rule of integration
Given: $f(x)=\int_{0}^{x} t \sin t\; d t$
Solution: $f(x)=\int_{0}^{x} t \sin t\; d t$
$=\left[t \int \sin t\; d t-\int\left\{\frac{d t}{d t} \int \sin t \; d t\right\} d t\right]_{0}^{x}$
\begin{aligned} &=\left[t(-\cos t)+\int \cos t d t\right]_{0}^{x} \\\\ &=[-t \cos t+\sin t]_{0}^{x} \end{aligned}
$=-x \cos x+\sin x-(-0 \cos 0+\sin 0)$
\begin{aligned} &f(x)=\sin x-x \cos x \\\\ &f^{\prime}(x)=\cos x-(x(-\sin x)+\cos x) \end{aligned}
\begin{aligned} &=\cos x+x \sin x-\cos x \\\\ &=x \sin x \end{aligned}

Definite Integrals exercise Very short answer type question 34

Answer: $a=2$
Hint: you must know the rule of integration
Given: $-\int_{0}^{a} \frac{1}{4+x^{2}} d x=\frac{\pi}{8}, \text { find } a$
Solution:
\begin{aligned} &\int_{0}^{a} \frac{1}{4+x^{2}} d x=\frac{\pi}{8} \\\\ &\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a} \end{aligned}
\begin{aligned} &\therefore \int_{0}^{a} \frac{1}{2^{2}+x^{2}} d x=\frac{\pi}{8} \\\\ &\Rightarrow \frac{1}{2} \tan ^{-1} \frac{a}{2}-0=\frac{\pi}{8} \end{aligned}
\begin{aligned} &\Rightarrow \tan ^{-1} \frac{a}{2}=\frac{\pi}{4} \\\\ &\Rightarrow \frac{a}{2}=\tan ^{-1} \frac{\pi}{4} \\\\ &\Rightarrow \frac{a}{2}=1 \end{aligned}
$\Rightarrow a=2$

Definite Integrals exercise Very short answer type question 35

Hint: Separate the terms of x and y and then integrate them.
Given:
Solution:
Integrating both sides

Definite Integrals exercise Very short answer type question 36

Answer: $\frac{18}{\log _{e} 3}$
Hint: you must know the rule of integration
Given: $I=\int_{2}^{3} 3^{x} d x$
Solution:
Let $y=3^{x}$ ..............(i)
Taking logarithm on both sides
\begin{aligned} &\log y=\log 3^{x} \\\\ &\log y=x \log 3 \\\\ &y=e^{x \log 3} \end{aligned} ........(ii)
From eqn (i) and (ii) we can write
$y=3^{x}=e^{x \log 3}$
This means instead of integrating $3^{x}$ we will integrate $e^{x \log 3}$
\begin{aligned} &I=\int_{2}^{3} 3^{x} d x \\\\ &=\int_{2}^{3} e^{x \log 3} d x \end{aligned}
\begin{aligned} &=\left[\frac{e^{x \log 3}}{\log 3}\right]_{2}^{3} \\\\ &=\frac{1}{\log 3}\left[e^{3 \log 3}-e^{2 \log 3}\right] \end{aligned}
\begin{aligned} &=\frac{1}{\log 3}\left[e^{\log 3^{3}}-e^{\log 3^{2}}\right] \\\\ &=\frac{1}{\log 3}\left[3^{3}-3^{2}\right] \end{aligned}\begin{aligned} &=\frac{(27-9)}{\log 3} \\\\ &=\frac{18}{\log 3} \end{aligned}

Definite Integrals exercise Very short answer type question 37

Hint: you must know the rule of integration
Given: $\int_{0}^{2}[x] d x$
Solution: $\int_{0}^{2}[x] d x$
\begin{aligned} &=\int_{0}^{1} 0 d x+\int_{1}^{2} 1 d x \\\\ &=0+[x]_{1}^{2} \\\\ &=2-1 \\\\ &=1 \end{aligned}

Definite Integrals exercise Very short answer type question 38

Answer: $\frac{1}{2}$
Hint: you must know the rule of integration
Given: $\int_{0}^{1.5}[x] d x$
Solution: $\int_{0}^{1.5}[x] d x$
\begin{aligned} &=\int_{0}^{1} 0 d x+\int_{1}^{1.5} 1 d x \\\\ &=[x]_{1}^{1.5} \end{aligned}
\begin{aligned} &=[1.5-1] \\\\ &=0.5 \\\\ &=\frac{1}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 39

Answer: $\frac{1}{2}$
Hint: you must know the rule of integration
Given: $\int_{0}^{1}\{x\} d x$
Solution: $\int_{0}^{1}\{x\} d x$
\begin{aligned} &=\int_{0}^{1}(x-[x]) d x \\\\ &=\int_{0}^{1} x d x-\int_{0}^{1}[x] d x \end{aligned}
\begin{aligned} =&\left[\frac{x^{2}}{2}\right]_{0}^{1}-\int_{0}^{1} 0 d x \\\\ =&\left[\frac{1}{2}-0\right]-0 \\ \end{aligned}
$=\frac{1}{2}$

Definite Integrals exercise Very short answer type question 40

Answer: $e-1$
Hint: you must know the rule of integration
Given: $\int_{0}^{1} e^{\{x\}} d x$
Solution: $\int_{0}^{1} e^{\{x\}} d x$
Integration of exponential function is same as the function
$\therefore\left[e^{\{x\}}\right]_{0}^{1} \quad\left[\because \int_{a}^{b} e^{x} d x=\left[e^{x}\right]_{a}^{b}\right]$
\begin{aligned} &= e^{1}-e^{0} \\\\ &=e-1 \end{aligned}

Definite Integrals exercise Very short answer type question 41

Answer: $\frac{3}{2}$
Hint: you must know the rule of integration
Given: $\int_{0}^{2} x[x] d x$
Solution: $\int_{0}^{2} x[x] d x$
$=\int_{0}^{1} x[0] d x+\int_{1}^{2} x(1) d x$
\begin{aligned} &=\left[0+\frac{x^{2}}{2}\right]_{1}^{2} \\\\ &=\frac{4}{2}-\frac{1}{2}=\frac{3}{2} \end{aligned}

Definite Integrals exercise Very short answer type question 42

Answer: $\frac{1}{\log _{e} 2}$
Hint: you must know the rule of integration
Given: $I=\int_{0}^{1} 2^{x-[x]} d x$
Solution:
We know the values of greatest integer function on $0
\begin{aligned} &I=\int_{0}^{1} 2^{x-0} d x \\\\ &=\int_{0}^{1} 2^{x} d x \\\\ &=\left[\frac{2^{x}}{\log 2}\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\left[\frac{2^{1}}{\log 2}-\frac{2^{0}}{\log 2}\right] \\\\ &=\frac{(2-1)}{\log 2} \\\\ &=\frac{1}{\log 2} \end{aligned}

### Definite Integrals exercise Very short answer type question 43

Hint: you must know the rule of integration
Given: $\int_{1}^{2} \log _{e}[x] d x$
Solution: $\int_{1}^{2} \log _{e}[x] d x$
Using rule of greatest integer
\begin{aligned} & \int_{0}^{1} 0 d x+\int_{1}^{2} 1 d x+\int_{2}^{3} 2 d x \\\\ I=& \int_{1}^{2} \log [x] d x=\int_{0}^{1} \log (0) d x+\int_{1}^{2} \log 1 d x \\\\ =& 0 \end{aligned}

Definite Integrals exercise Very short answer type question 44

Answer: $\sqrt{2}-1$
Hint: The values of greatest integer function when $0
And when $1
Given: $\int_{0}^{\sqrt{2}}\left[x^{2}\right] d x$
Solution: $\int_{0}^{\sqrt{2}}\left[x^{2}\right] d x$
$=\int_{0}^{1}\left[x^{2}\right] d x+\int_{1}^{\sqrt{2}}\left[x^{2}\right] d x$ [ using the value of greatest integer function ]
\begin{aligned} &=0+\int_{1}^{\sqrt{2}} 1 d x \\\\ &=[x]_{1}^{\sqrt{2}} \\\\ &=\sqrt{2}-1 \end{aligned}
Note: Final answer is not matching with the book.

Definite Integrals exercise Very short answer type question 45

Answer: $\frac{\sqrt{2}-1}{\sqrt{2}}$
Hint: you must know the rule of integration
Given: $\int_{0}^{\frac{\pi}{4}} \sin \{x\} d x$
Solution: $\int_{0}^{\frac{\pi}{4}} \sin \{x\} d x$
\begin{aligned} &=-[\cos \{x\}]_{0}^{\frac{\pi}{4}} \\\\ &=-\left[\cos \frac{\pi}{4}-\cos 0\right] \end{aligned}
\begin{aligned} &=-\left[\frac{1}{\sqrt{2}}-1\right] \\\\ &=\frac{\sqrt{2}-1}{\sqrt{2}} \end{aligned}

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## RD Sharma Chapter-wise Solutions

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1. What are definite integrals?

The Definite integral is defined to be exactly the same limit and the sum that we see between any function and the X-axis.

2. What does a definite integral represent?

A definite integral represents a number only  when the lower and upper limits of the integral are constant.  The Definite integral is the representation of a function family whose derivatives are f.

3. What is the importance of RD Sharma class 12 solutions chapter 19 exercise FBQ?

A definite integral chapter helps you understand the constant and functions when it comes to understanding of the antiderivatives.

4. Is the  definite integral more important than the indefinite integral?

Both the chapters of mathematics are important for every student to practice,  therefore it can't be said which of the chapters will be more important for you to prepare.

5. How many questions are there in this exercise?

There are a total of 9 questions in this exercise which will help you to practice  the fill in the blank type of questions easily.

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