RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA

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RD Sharma Class 12 Solutions Chapter19 VSA Definite Integrals - Other Exercise
Definite Integrals Excercise:VSA
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter19 VSA Definite Integrals - Other Exercise

Definite Integrals Excercise:VSA

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

Answer: 0
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

Answer: 0
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: $\int_{-2}^{1} \frac{|x|}{x} d x$
$\begin{aligned} &=\int_{-2}^{0} \frac{-x}{x} d x+\int_{0}^{1} \frac{x}{x} d x \\\\ &=\int_{-2}^{0}-1 d x+\int_{0}^{1} 1 d x \end{aligned}$
$\begin{aligned} &=-[x]_{-2}^{0}+[x]_{0}^{1} \\\\ &=-[0+2]+[1+0] \\\\ &=-2+1 \\\\ &=-1 \end{aligned}$

Definite Integrals exercise Very short answer type question 10

Answer: 1
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

Answer: 0
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)
Adding (i) and (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

Answer: 0
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)
Adding (i) and (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)
Add (i) and (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

Answer: 0
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

Answer: 0
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

Answer: 0
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)
Add (i) and (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

Answer: 1
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

Answer: 2
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

Answer:
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

Answer: 1
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 $0<x<1, \quad[x]=0$
$\begin{aligned} &I=\int_{0}^{1} 2^{x-0} d x \\\\ &=\int_{0}^{1} 2^{x} d x \\\\ &=\left[\frac{2^{x}}{\log 2}\right]_{0}^{1} \end{aligned}$
$\begin{aligned} &=\left[\frac{2^{1}}{\log 2}-\frac{2^{0}}{\log 2}\right] \\\\ &=\frac{(2-1)}{\log 2} \\\\ &=\frac{1}{\log 2} \end{aligned}$

Definite Integrals exercise Very short answer type question 43

Answer: 0
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 $0<x<1,[x]=0$
And when $1<x<2,[x]=1$
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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