RD Sharma Class 12 Exercise 19.3 Definite Integrals Solutions Maths

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Chapter 19 - Definite Integrals - Ex-19.1

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Chapter 19 - Definite Integrals - Ex-MCQ

Definite Integrals Exercise 19.3 Question 1(i)

Answer: $37$
Hint: Break the range of integration and then solve it.
Given:
$\int_{1}^{4} f(x) d x \text { where } f(x)=\left\{\begin{array}{ll} 4 x+3, & \text { if } 1 \leq x \leq 2 \\ 3 x+5, & \text { if } 2 \leq x \leq 4 \end{array}\right\}$
Solution:
$\int_{1}^{4} f(x) d x$
Now break the limit = $\int_{1}^{2} f(x) d x+\int_{2}^{4} f(x) d x=\int_{1}^{2}(4 x+3) d x+\int_{2}^{4}(3 x+5) d x$
$=\left[\frac{4 x^{2}}{2}+3 x\right]_{1}^{2}+\left[\frac{3 x^{2}}{2}+5 x\right]_{2}^{4}$ $\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\left(2 x^{2}+3 x\right)_{1}^{2}+\left(\frac{3 x^{2}}{2}+5 x\right)_{2}^{4} \\ & \end{aligned}$
$=2(4-1)+3(2-1)+\frac{3}{2}(16-4)+5(4-2) \\$
$=6+3+18+10 \\$
$=37$

Definite Integrals Exercise 19.3 Question 1(ii)

Answer: $3-\frac{\pi}{2}+e^{6}$
Hint: Break the range of integration from $0 \: to \: \frac{\pi}{2},\: \frac{\pi}{2} \: to\: 3,$ and then $3\: to \: 9$
Given:
$\int_{0}^{9} f(x) d x \text { where } f(x)=\left\{\begin{array}{ll} \sin x, & \text { if } 0 \leq x \leq \frac{\pi}{2} \\\\ 1, & \text { if } \frac{\pi}{2} \leq x \leq 3 \\\\ e^{x-3}, & \text { if } 3 \leq x \leq 9 \end{array}\right\}$
Solution:
$\int_{0}^{9} f(x) d x$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} f(x) d x+\int_{\frac{\pi}{2}}^{3} f(x) d x+\int_{3}^{9} f(x) d x \\ & \end{aligned}$
$=\int_{0}^{\frac{\pi}{2}} \sin x d x+\int_{\frac{\pi}{2}}^{3} 1 d x+\int_{3}^{9} e^{x-3} d x \\$
$=[-\cos x]_{0}^{\frac{\pi}{2}}+[x]_{\frac{\pi}{2}}^{3}+\left[\frac{e^{x}}{e^{3}}\right]_{3}^{9}$
$\begin{aligned} &=-\cos \frac{\pi}{2}+\cos 0+3-\frac{\pi}{2}+\frac{e^{9}-e^{3}}{e^{3}} \\ & \end{aligned}$
$=0+1+3-\frac{\pi}{2}+\frac{e^{3}\left(e^{6}-1\right)}{e^{3}} \\$
$=4-\frac{\pi}{2}+e^{6}-1 \\$
$=3-\frac{\pi}{2}+e^{6}$

Definite Integrals Exercise 19.3 Question 1(iii)

Answer: $62$
Hint: Break the range of integration and then solve the integration.
Given: $\int_{1}^{4} f(x) d x \text { where } f(x)$
$=\left\{\begin{array}{ll} 7 x+3, & \text { if } 1 \leq x \leq 3 \\ 8 x, & \text { if } 3 \leq x \leq 4 \end{array}\right\}$
Solution:
$\int_{1}^{4} f(x) d x$
$\begin{aligned} &=\int_{1}^{3} f(x) d x+\int_{3}^{4} f(x) d x \\ & \end{aligned}$
$=\int_{1}^{3}(7 x+3) d x+\int_{3}^{4} 8 x d x$ $\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\left[\frac{7 x^{2}}{2}+3 x\right]_{1}^{3}+\left[\frac{8 x^{2}}{2}\right]_{3}^{4} \\ & \end{aligned}$
$=\left[7 \frac{(9-1)}{2}+3(3-1)\right]+4(16-9) \\$
$=7(4)+3(2)+4(7) \\$
$=28+6+28 \\$
$=62$

Definite Integrals Exercise 19.3 Question 1(iv)

Answer: $1$
Hint: Break the range of integration $\left\{\begin{array}{c} -1<x<0 \\ 0<x<2 \end{array}\right\}$
Given: $\int_{-1}^{2} \frac{|x|}{x}$
Solution:
$\begin{aligned} &|x|=\left\{\begin{array}{l} -x, \text { if }-1 \leq x \leq 0 \\ x, \text { if } 0 \leq x \leq 2 \end{array}\right\} \\ \end{aligned}$
$\frac{|x|}{x}=\left\{\begin{array}{l} -1,-1<x<0 \\ 1,0<x<2 \end{array}\right\} \\$
$I=\int_{-1}^{0}(-1) d x+\int_{0}^{2}(1) d x$

Definite Integrals Excercise 19.3 Question 3

Answer: $10$
Hint: Break the range of integration like this $\int_{-3}^{-1} f(x) \& \int_{-1}^{3} f(x)$
Given: $\int_{-3}^{3}|x+1| d x$
Solution:
$\begin{aligned} &I=\int_{-3}^{3}|x+1| d x \\ & \end{aligned}$
$f(x)=|x+1|=\left\{\begin{array}{ll} -(x+1), & \text { if }-3 \leq x \leq-1 \\ (x+1), & \text { if }-1 \leq x \leq 3 \end{array}\right\}$
$\begin{aligned} &I=\int_{-3}^{-1} f(x) d x+\int_{-1}^{3} f(x) d x \\ & \end{aligned}$
$I=\int_{-3}^{-1}-(x+1) d x+\int_{-1}^{3}(x+1) d x$
$\begin{aligned} &I=\left[-\left(\frac{x^{2}}{2}+x\right)\right]_{-3}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{3} \\ & \end{aligned}$ $\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$y=-\left[\frac{(1-9)}{2}+(-1+3)\right]+\frac{(9-1)}{2}+3+1$
$\begin{aligned} &=-(-4+2)+(4+4) \\ & \end{aligned}$
$=-(-2)+8 \\$
$=2+8 \\$
$=10$

Definite Integrals Excercise 19.3 Question 4

Answer: $\frac{5}{2}$
Hint: Break the range of integration and then integrate.
Given: $\int_{-1}^{1}|2 x+1| d x$
Solution:
$f(x)=2 x+1=\left\{\begin{array}{ll} -(2 x+1), & \text { if }-1 \leq x \leq \frac{-1}{2} \\ (2 x+1), & \text { if } \frac{-1}{2} \leq x \leq 1 \end{array}\right\}$
$\begin{aligned} &I=\int_{-1}^{\frac{-1}{2}} f(x) d x+\int_{\frac{-1}{2}}^{1} f(x) d x \\ & \end{aligned}$
$I=\int_{-1}^{\frac{-1}{2}}-(2 x+1) d x+\int_{\frac{-1}{2}}^{1}(2 x+1) d x$
Using the formula: $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &I=\left[-\left(\frac{2 x^{2}}{2}+x\right)\right]_{-1}^{-\frac{1}{2}}+\left[\frac{2 x^{2}}{2}+x\right]_{\frac{-1}{2}}^{1} \\ & \end{aligned}$
$I=-\left[\left(x^{2}+x\right)\right]_{-1}^{\frac{-1}{2}}+\left[x^{2}+x\right]_{\frac{-1}{2}}^{1}$
$\begin{aligned} &=-\left(\left(\frac{-1}{2}\right)^{2}+\left(\frac{-1}{2}\right)-1+1\right)+\left(1+1-\frac{1}{4}+\frac{1}{2}\right) \\ & \end{aligned}$
$=\frac{-1}{4}+\frac{1}{2}+2-\frac{1}{4}+\frac{1}{2} \\$
$=\frac{-2}{4}+\frac{2}{2}+2 \\$
$=\frac{5}{2}$

Definite Integrals Excercise 19.3 Question 5

Answer: $\frac{25}{2}$
Hint: Break the range of integration and then solve the integration.
Given: $\int_{-2}^{2}|2 x+3| d x$
Solution:
$f(x)=|2 x+3|$
$f(x)=\left\{\begin{array}{ll} -(2 x+3), & \text { if }-2 \leq x \leq \frac{-3}{2} \\ (2 x+3), & \text { if } \frac{-3}{2} \leq x \leq 2 \end{array}\right\}$ $\quad\left(\begin{array}{c} 2 x+3=0 \\ x=\frac{-3}{2} \end{array}\right)$
$\begin{aligned} &I=\int_{-2}^{\frac{-3}{2}} f(x) d x+\int_{\frac{-3}{2}}^{2} f(x) d x \\ & \end{aligned}$
$I=\int_{-2}^{\frac{-3}{2}}-(2 x+3) d x+\int_{\frac{-3}{2}}^{2}(2 x+3) d x$
Use the formula: $\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} I &=\left[-\left(x^{2}+3 x\right)\right]_{-2}^{-\frac{3}{2}}+\left[x^{2}+3 x\right]_{\frac{-3}{2}}^{2} \\ \end{aligned}$
$=-\left[\left(\frac{-3}{2}\right)^{2}-(-2)^{2}\right]-3\left(\frac{-3}{2}+2\right)+\left[(2)^{2}-\left(-\frac{3}{2}\right)^{2}+3\left(2+\frac{3}{2}\right)\right] \\$
$=-\left(\frac{9}{4}-4-\frac{9}{2}+6\right)+\left(4-\frac{9}{4}+6+\frac{9}{2}\right) \\$
$= \frac{25}{2}$

Definite Integrals Exercise 19.3 Question 6

Answer: $1$
Hint: Use distribution method to find the value of x.
Given: $\int_{0}^{2}\left|x^{2}-3 x+2\right| d x$
Solution:
$\int_{0}^{2}\left|x^{2}-3 x+2\right| d x$ $\left[\left(x^{2}+3 x+2\right)=(x-2)(x-1)\right]$
Also we know that:
$|x|=\left\{\begin{array}{l} x, x \geq 0 \\ -x, x \leq 0 \end{array}\right\}$
$\int_{0}^{1}\left(x^{2}-3 x+2\right) d x-\int_{1}^{2}\left(x^{2}-3 x+2\right) d x$
$\begin{aligned} &=\left[\frac{x^{3}}{3}-\frac{3 x^{2}}{2}+2 x\right]_{0}^{1}-\left[\frac{x^{3}}{3}-\frac{3 x^{2}}{2}+2 x\right]_{1}^{2} \\ \end{aligned}$
$=\frac{1}{3}-\frac{3}{2}+2-\left[\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\right] \\$
$=\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-2+\frac{1}{3}-\frac{3}{2} \\$
$=1$

Definite Integrals Exercise 19.3 Question 7

Answer: $\frac{13}{2}$
Hint: Use $x=\frac{1}{2}$ ,
$\begin{aligned} &2 x-1=0 \\ & \end{aligned}$
$x=\frac{1}{2}$
Given: $\int_{0}^{3}|2 x-1| d x$
$I=\int_{0}^{3} f(x) d x$
$f(x)=\left\{\begin{array}{ll} -(2 x-1), & \text { if } 0 \leq x \leq \frac{1}{2} \\ (2 x-1), & \text { if } \frac{1}{2} \leq x \leq 3 \end{array}\right\}$
$=\int_{0}^{\frac{1}{2}}-(2 x-1) d x+\int_{\frac{1}{2}}^{3}(2 x-1) d x$
$I=\left[-\left(x^{2}-x\right)\right]_{0}^{\frac{1}{2}}+\left(x^{2}-x\right)_{\frac{1}{2}}^{3}$
$\begin{aligned} I &=-\left[\left(\frac{1}{2}\right)^{2}-\frac{1}{2}-0\right]+\left[(3)^{2}-3-\left(\left(\frac{1}{2}\right)^{2}-\frac{1}{2}\right)\right] \\ \end{aligned}$
$=\frac{-1}{4}+\frac{1}{2}+6-\frac{1}{4}+\frac{1}{2} \\$
$=\frac{-1}{2}+1+6 \\$
$=\frac{-1+14}{2} \\$
$=\frac{13}{2}$

Definite Integrals Exercise 19.3 Question 8

Answer: $40$
Hint: Break the range of integration and then solve.
Given: $\begin{aligned} &\int_{-6}^{6}|x+2| d x \\ & \end{aligned}$
$x+2=0, x=-2$
Solution:
$\begin{aligned} \int_{-6}^{6}|x+2| d x & \\ \end{aligned}$
$\left\{\begin{array}{ll} -(x+2), & -6 \leq x \leq-2 \\ (x+2), & -2 \leq x \leq-6 \end{array}\right\}$
$\begin{aligned} &=\int_{-6}^{-2}-(x+2) d x+\int_{-2}^{6}(x+2) d x \\ & \end{aligned}$
$=\left[-\left(\frac{x^{2}}{2}+2 x\right)\right]_{-6}^{-2}+\left(\frac{x^{2}}{2}+2 x\right)_{-2}^{6} \\$
$=-2+4+18-12+18+12-2+4=40$

Definite Integrals Exercise 19.3 Question 9

Answer: $5$
Hint: You must know the rules of solving definite integral.
Given:
$\begin{aligned} &\int_{-2}^{2}|x+1| d x \\ \end{aligned}$
$x+1=0 \\$
$x=-1$
Solution:
$\begin{aligned} &\int_{-2}^{2}|x+1| d x \\ \end{aligned}$
$\left\{\begin{array}{c} -(x+1),-2 \leq x \leq-1 \\ x+1, \quad-1 \leq x \leq 2 \end{array}\right\}$
$\begin{aligned} &=\int_{-2}^{-1}-(x+1) d x+\int_{-1}^{2}(x+1) d x \\ & \end{aligned}$
$=\left[-\left(\frac{x^{2}}{2}+x\right)\right]_{-2}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{2}$
$\begin{aligned} &=\frac{-1}{2}+1+2-2+2+2-\frac{1}{2}+1 \\ & \end{aligned}$
$=5$

Definite Integrals Exercise 19.3 question 10

Answer:

Hint: You must know the rules of solving definite integral.
Given: $\begin{aligned} &\int_{1}^{2}|x-3| d x \\ & \end{aligned}$
$x-3=0 \\$
$x=3$
Solution:
$\begin{aligned} \mathrm{I} &=\int_{1}^{2}|x-3| d x \\ & \end{aligned}$
$=\int_{1}^{2}-(x-3) d x \\$
$=\left[-\left(\frac{x^{2}}{2}-3 x\right)\right]_{1}^{2}$
$\begin{aligned} &l=-\left(\frac{4}{2}-6-\frac{1}{2}+3\right) \\ & \end{aligned}$
$I=-\left(\frac{4}{2}-6-\frac{1}{2}+3\right) \\$
$I=-\left(\frac{3}{2}-3\right)$
$\begin{aligned} I &=-\frac{3}{2}+\frac{3}{1} \\ \end{aligned}$
$=\frac{-3+6}{2} \\$
$=\frac{3}{2}$

Definite Integrals Exercise 19.3 question 11

Answer: $0$
Hint: You must know the rules of solving definite integral.
Given: $\int_{0}^{\frac{\pi}{2}}|\cos 2 x| d x$
Solution:
$\int_{0}^{\frac{\pi}{2}}|\cos 2 x| d x$
We know that $|\cos 2 x|=\left\{\begin{array}{ll} -\cos 2 x & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\\\ \cos 2 x & 0<x \leq \frac{\pi}{4} \end{array}\right\}$
$\begin{aligned} &\therefore I=\int_{0}^{\frac{\pi}{2}}|\cos 2 x| d x \\ & \end{aligned}$
$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos 2 x d x$
$\begin{aligned} &\Rightarrow I=\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{4}}-\left[\frac{\sin 2 x}{2}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{2}-0-0+\frac{1}{2} \\$
$\Rightarrow I=1$

Definite Integrals Exercise 19.3 question 12

Answer: $4$
Hint: You must know the rules of solving definite integral.
Given: $\int_{0}^{2 \pi}|\sin x| d x$

Solution:
$\begin{aligned} &I=\int_{0}^{2 \pi}|\sin x| d x=\int_{0}^{\pi} \sin x d x-\int_{\pi}^{2 \pi} \sin x d x \\ & \end{aligned}$
$I=\int_{0}^{\pi} \sin x d x-\int_{\pi}^{2 \pi} \sin x d x \\$
$I=(-\cos x)_{0}^{\pi}-(-\cos x)_{\pi}^{2 \pi} \\$
$I=(1+1)-(-1-1) \\$
$I=2-(-2) \\$
$I=4$

Definite Integrals Exercise 19.3 question 13

Answer: $2-\sqrt{2}$
Hint: You must know about the rules of solving definite integral.
Given: $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}|\sin x| d x$
Solution:
$\mathrm{I}=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}|\sin x| d x$

$\begin{gathered} I=\int_{\frac{-\pi}{4}}^{0}(-\sin x) d x-\int_{0}^{\frac{\pi}{4}} \sin x d x \\ \end{gathered}$
$\begin{gathered} \left\{\begin{array}{l} \sin x>0,\left[0, \frac{\pi}{4}\right] \\\\ \sin x<0\left[-\frac{\pi}{4}, 0\right] \end{array}\right\} \end{gathered}$

$\begin{gathered} I=(\cos x) \frac{0}{\frac{0 \pi}{4}}+(\cos x)_{0}^{\frac{\pi}{4}} \\ \end{gathered}$
$I=1-\cos \left(\frac{-\pi}{4}\right)+\left(\cos \frac{\pi}{4}-\cos 0\right) \\$
$I=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1 \\$
$I=2-\sqrt{2}$

Definite Integrals Exercise 19.3 Question 14

Answer: $\frac{15}{2}$
Hint: Use negative value of x.

Given:
$\int_{1}^{4}|x-5| d x$
Solution:
$\begin{aligned} I=\int_{1}^{4}|x-5| d x \\ \end{aligned}$
$\left\{\begin{array}{l} x-5, x \geq 5 \\ -(x-5), x<5 \end{array}\right\}$
$\begin{aligned} &I=\int_{1}^{4}(-x+5) d x \\ & \end{aligned}$
$=\left(\frac{-x^{2}}{2}+5 x\right)_{1}^{4} \\$
$=12-\frac{9}{2} \\$
$=\frac{15}{2}$

Definite Integrals Exercise 19.3 Question 15

Answer: $4$
Hint: We will use the concept of even function to solve the integral.
Given:
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\{\sin |x|+\cos |x|\} d x$
Solution:
$\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\{\sin |x|+\cos |x|\} d x \\ & \end{aligned}$
$f(-x)=\sin |-x|+\cos |-x|=\sin |x|+\cos |x|=f(x)$

$\begin{aligned} &f(x) \text { is a even function }\\ & \end{aligned}$
$I=2 \int_{0}^{\frac{\pi}{2}}(\sin x+\cos x) d x$
$\begin{aligned} &{\left[\text { for even function } \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x d x=2 \int_{0}^{\frac{\pi}{2}} x d x\right]} \\ & \end{aligned}$
$I=2[-\cos x+\sin x]_{0}^{\frac{\pi}{2}} \\$
$I=2[(0+1)-(-1-0)] \\$
$I=2(2)=4$

Definite Integrals Exercise 19.3 Question 16

Answer: $5$
Hint: You must know the rules of solving definite integral.
Given:
$\int_{0}^{4}|x-1| d x$
Solution:
$\begin{aligned} &\mathrm{I}=\int_{0}^{4}|x-1| d x \\ & \end{aligned}$
$|x-1|=\left\{\begin{array}{l} -(x-1), 0 \leq x \leq 1 \\ x-1, \quad 1 \leq x \leq 4 \end{array}\right\}$
$\begin{aligned} I &=\int_{0}^{1}-(x-1) d x+\int_{1}^{4}(x-1) d x \\ & \end{aligned}$
$=\left[-\left(\frac{x^{2}}{2}-x\right)\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4} \\$
$=\frac{1}{2}+8-4+\frac{1}{2}=4+1 \\$
$I =5$

Definite Integrals Exercise 19.3 Question 17

Answer: $\frac{23}{2}$
Hint: You must know about the rules of solving definite integral.
Given: $\int_{1}^{4}\{|x-1|+|x-2|+|x-4|\} d x$
Solution:
$\begin{aligned} &I=\int_{1}^{4}\{|x-1|+|x-2|+|x-4|\} d x \\ \end{aligned}$
$\mathrm{I}=\int_{1}^{4}|x-1| d x+\int_{1}^{4}|x-2| d x+\int_{1}^{4}|x-4| d x$
We know that,
$\begin{aligned} &|x-1|=\left\{\begin{array}{l} -(x-1), x \leq 1 \\ x-1,1<x \leq 4 \end{array}\right\} \\ \end{aligned}$
$|x-2|=\left\{\begin{array}{l} -(x-2), 1 \leq x \leq 2 \\ x-2,2 \leq x \leq 4 \end{array}\right\} \\$
$|x-3|=\left\{\begin{array}{l} -(x-4), 1 \leq x \leq 4 \\ x-4, x>4 \end{array}\right\}$
$\int_{1}^{4}(x-1)+\int_{1}^{2}-(x-2) d x+\int_{2}^{4}(x-2) d x+\int_{1}^{4}-(x-4) d x$

$\begin{aligned} &=\left(\frac{x^{2}}{2}-x\right)_{1}^{4}+\left(\frac{-x^{2}}{2}+2 x\right)_{1}^{2}+\left(\frac{x^{2}}{2}-2 x\right)_{2}^{4}+\left(\frac{-x^{2}}{2}+4 x\right)_{1}^{4} \\ \end{aligned}$
$=\left(\frac{16}{2}-4-\frac{1}{2}+1\right)+\left(-2+4+\frac{1}{2}-2\right)+\left(\frac{16}{2}-8-2+4\right)+\left(-\frac{16}{2}+16+\frac{1}{2}-4\right) \\$
$=\frac{23}{2}$

Definite Integrals Excercise 19.3 Question 18

Answer: $\frac{63}{2}$
Hint: You must know the rules of solving definite integral.
Given: $\int_{-5}^{0} f(x) d x \text { where } f(x)=|x|+|x+2|+|x+5|$
Solution:
For the first integrand:
$\begin{aligned} &x<0 \\ & \end{aligned}$
$|x|=-x$
$\begin{aligned} \int_{-5}^{0}|x| d x & \\ & \end{aligned}$
$=\int_{-5}^{0}-x d x \\$
$=-\left[\frac{x^{2}}{2}\right]_{-5}^{0} \\$
$=\frac{25}{2}$
For the second integrand:
$\begin{gathered} (x+2)=\left\{\begin{array}{l} x+2, \text { where } x \geq-2 \\\\ -(x+2), \text { wherex } \leq-2 \end{array}\right\} \\ \end{gathered}$
$\int_{-5}^{0}|x+2| d x=\int_{-5}^{-2}-(x+2) d x+\int_{-2}^{0}(x+2) d x$
For the third integrand:
$\begin{array}{r} |x+5|=x+5, \text { if } x \geq-5 \\ \end{array}$
$\int_{-5}^{0}|x+5| d x=\int_{-5}^{0}(x+5) d x$
$\begin{aligned} &=\left(\frac{x^{2}}{2}+5 x\right)_{-5}^{0} \\ & \end{aligned}$
$=0-\frac{25}{2}+25 \\$
$=\frac{25}{2}$
Hence the total integration will be
$\begin{aligned} \int_{-5}^{0} f(x) d x &=\int_{-5}^{0}|x| d x+\int_{-5}^{0}|x+2| d x+\int_{-5}^{0}|x+5| d x \\ & \end{aligned}$
$=\frac{25}{2}+\frac{13}{2}+\frac{25}{2} \\\\$
$=\frac{63}{2}$

Definite Integrals Excercise 19.3 Question 19

Answer: $20$
Hint: You must know the rules of solving definite integral.
Given: $\int_{0}^{4}(|x|+|x-2|+|x-4|) d x$
Solution:
$\begin{aligned} &I=\int_{0}^{4}(|x|+|x-2|+|x-4|) d x \\ & \end{aligned}$
$\Rightarrow I=\int_{0}^{4}|x| d x+\int_{0}^{4}|x-2| d x+\int_{0}^{4}|x-4| d x$
We know that,
$\begin{aligned} &|x|= \begin{cases}-x & -5 \leq x \leq 0 \\ x & x>0\end{cases} \\ & \end{aligned}$
$|x-2|= \begin{cases}-(x-2) & 0 \leq x \leq 2 \\ x-2 & 2<x \leq 4\end{cases}$
$\begin{aligned} &|x-4|=\left\{\begin{array}{lc} -(x-4) & 0 \leq x \leq 4 \\ x-4 & x>4 \end{array}\right. \\ & \end{aligned}$
$\therefore I=\int_{0}^{4} x d x-\int_{0}^{4}(x-2) d x+\int_{0}^{4}(x-2) d x-\int_{0}^{4}(x-4) d x$
$\begin{aligned} &\Rightarrow I=\left[\frac{x^{2}}{2}\right]_{0}^{4}-\left[\frac{x^{2}}{2}-2 x\right]_{0}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{4}-\left[\frac{x^{2}}{2}-4 x\right]_{0}^{4} \\ & \end{aligned}$
$\Rightarrow I=8-(2-4)+8-8-2+4-(8-16) \\$
$\Rightarrow I=20$

Definite Integrals Excercise 19.3 Question 20

Answer: $\frac{19}{2}$
Hint: You must know the rules of solving definite integral.
Given: $\int_{-1}^{2}|x+1|+|x|+|x-1| d x$
Solution:
$\begin{aligned} &f(x)=|x+1|+|x|+|x-1| \\ & \end{aligned}$
$f(x)=x+1-x-x+1=-x+2 \\$
$f(x)=x+1+x-x+1 \rightarrow 0 \leq x \leq 1=2+x$
$\begin{aligned} &f(x)=x+1+x+x-1 \rightarrow x \geq 1=3 x \\ & \end{aligned}$
$f(x)=-x-1-x-x+13-3 x=0$
$\begin{aligned} &=\int_{-1}^{0}(2-x) d x+\int_{0}^{1}(x+2) d x+\int_{1}^{2} 3 x d x \\ & \end{aligned}$
$=\left[2 x-\frac{x^{2}}{2}\right]_{-1}^{0}+\left(\frac{x^{2}}{2}+2 x\right)_{0}^{1}+\left(\frac{3 x^{2}}{2}\right)_{1}^{2}$
$\begin{aligned} &=2+\frac{1}{2}+\frac{1}{2}+2+6-\frac{3}{2} \\ & \end{aligned}$
$=\frac{19}{2}$

Definite Integrals Excercise 19.3 Question 21

Answer: $0$
Hint: Use ILATE ,( Inverse , Logarithm , Algebraic , Trigonometric , Exponent.)
Given:
$\int_{-2}^{2} x e^{|x|} d x$
Solution:
Consider
$f(x)=x e^{|x|}$
Now
$\begin{aligned} &f(-x)=(-x) e^{|-x|}=-x e^{|x|}=-f(x) \\ \end{aligned}$
$\therefore \int_{-2}^{2} x e^{|x|} d x=0 \\$
${\left[\int_{-a}^{a} f(x) d x=\left\{\begin{array}{ll} 2 \int_{0}^{a} f(x) d x & \text { if } f(-x)=f(x) \\ 0 & \text { if } f(-x)=-f(x) \end{array}\right]\right.}$

Definite Integrals Excercise 19.3 Question 22

Answer: $\frac{\pi}{8}+\frac{1}{4}$
Hint: You must know the rules of solving definite integral.
Given: $\int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} \sin x|\sin x| d x$
Solution:
$\begin{aligned} &\mathrm{I}=\int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} \sin x|\sin x| d x \\ & \end{aligned}$
$I=-\int_{\frac{-\pi}{4}}^{0} \sin ^{2} x d x+\int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$
We will use the formula
So,
$\begin{aligned} &x=\frac{1-\cos 2 x}{2} \\ & \end{aligned}$
$=-\int_{\frac{-\pi}{4}}^{0} \frac{1-\cos 2 x}{2} d x+\int_{0}^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) d x$
$\begin{aligned} &=-\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)_{\frac{-\pi}{4}}^{0}+\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)_{0}^{\frac{\pi}{2}} \\ \end{aligned}$
$=-\frac{1}{2}\left(0-\left(-\frac{\pi}{4}+\frac{1}{2}\right)+\frac{1}{2}\left(\frac{\pi}{2}-0\right)\right. \\$
$=-\frac{\pi}{8}+\frac{1}{4}+\frac{\pi}{4} \\$
$=\frac{\pi}{8}+\frac{1}{4}$

Definite Integrals Excercise 19.3 Question 23

Answer: $0$
Hint: We will use the property of definite integrals.
Given: $\int_{0}^{\pi} \cos x|\cos x| d x$
Solution:
$I=\int_{0}^{\pi} \cos x|\cos x| d x$ … (i)
Consider, $f(x)=\cos x|\cos x|$
Now, use the property: $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$
$I=\int_{0}^{\pi} \cos (\pi-x)|\cos (\pi-x)| d x \\$
$=\int_{0}^{\pi}-\cos (x)|\cos (x)| d x \\$
$\begin{aligned} & &I=\int_{0}^{\pi}-\cos (x)|\cos (x)| d x \end{aligned}$ … (ii)
Adding (i) and (ii) , we get
$\begin{aligned} &2 I=0 \\ & \end{aligned}$
$I=0$

Definite Integrals Exercise 19.3 Question 24

Answer: $6$
Hint: We will check for the nature of function ( even or odd) then will use the property of definition
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x$
Given: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x$
Solution:
$\begin{aligned} &\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(2 \sin |\mathrm{x}|+\cos |\mathrm{x}|) \mathrm{d} x \\ \end{aligned}$
$\mathrm{f}(\mathrm{x})=2 \sin |\mathrm{x}|+\cos |\mathrm{x}| \\$
$\begin{aligned} \mathrm{f}(-\mathrm{x})=2 \sin |-\mathrm{x}|+\cos |-\mathrm{x}| \\ \end{aligned}$
$=2 \sin |\mathrm{x}|+\cos |\mathrm{x}|=\mathrm{f}(\mathrm{x})$
This shows that f(x) is an even function.
So we use the property :
$\begin{aligned} &\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x \\ & \end{aligned}$
$\text { if } f(-x)=f(x)$
$\begin{aligned} &I=2 \int_{0}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x \\ \end{aligned}$
$I=2 \int_{0}^{\frac{\pi}{2}}(2 \sin x+\cos x) d x$
$\begin{aligned} &=-2(\cos x \times 2)_{0}^{\frac{\pi}{2}}+2(\sin x)_{0}^{\frac{\pi}{2}} \\ & \end{aligned}$
$=-4(0-1)+2(1-0)=(4+2)=6$

Definite Integrals Exercise 19.3 Question 25

Answer: $\frac{\pi^{2}}{8}$
Hint: You must know the rules of solving definite integral.
Given:
$\int_{\frac{-\pi}{2}}^{\pi} \sin ^{-1}(\sin x) d x$
Solution:
$\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{-1}(\sin x) d x \\ \end{aligned}$ `
$\sin ^{-1}(\sin x)=\left\{\begin{array}{l} x, \quad \frac{-\pi}{2} \leq x \leq \frac{\pi}{2} \\\\ (\pi-x), \frac{\pi}{2} \leq x \leq \frac{3 \pi}{2} \end{array}\right\}$
$\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{-1}(\sin x) d x+\int_{\frac{\pi}{2}}^{\pi} \sin ^{-1}(\sin x) d \\ & \end{aligned}$
$I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x d x+\int_{\frac{\pi}{2}}^{\pi}(\pi-x) d x$ $\quad\left[\sin ^{-1}(\sin x)=x\right]$
$\begin{aligned} &I=\left(\frac{x^{2}}{2}\right)_{\frac{-\pi}{2}}^{\frac{\pi}{2}}+\left(\pi x-\frac{x^{2}}{2}\right)_{\frac{\pi}{2}}^{\pi} \\ \end{aligned}$
$I=\frac{1}{2}\left(\frac{\pi^{2}}{4}-\frac{\pi^{2}}{4}\right)+\left(\pi \times \pi-\frac{\pi^{2}}{2}-\left(\pi \times \frac{\pi}{2}-\frac{\frac{\pi^{2}}{4}}{2}\right)\right) \\$
$=\pi^{2}-\frac{\pi^{2}}{2}-\frac{\pi^{2}}{2}+\frac{\pi^{2}}{8}=\frac{\pi^{2}}{8}$

Definite Integrals Exercise 19.3 Question 26

Answer: $-\infty$
Given: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{-\frac{\pi}{2}}{\sqrt{\cos x \sin ^{2} x}} d x$
Solution:
Let
$\begin{aligned} &I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{-\frac{\pi}{2}}{\sqrt{\cos x \sin ^{2} x}} d x \\ & \end{aligned}$
$=-\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x \sin ^{2} x}} d x$
$\begin{aligned} &=-\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x}|\sin x|} d x\\ \end{aligned}$
$=-\frac{\pi}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x}|\sin x|} d x\\$
$\begin{aligned} &|\sin x|=\sin x, 0 \leq x \leq \frac{\pi}{2} \\ & \end{aligned}$
$=-\pi \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x} \sin x} d x \\$
$=-\pi \int_{0}^{\frac{\pi}{3}} \frac{1}{\sqrt{\cos x}\left(1-\cos ^{2} x\right)} d x$
Put $\cos x=t^{2}$ and differentiate on both sides
$\begin{aligned} &-\sin x d x=2 t d t \\ & \end{aligned}$
$\text { when } x \rightarrow 0, t \rightarrow 1 \\$
$\text { when } x \rightarrow \frac{\pi}{2}, t \rightarrow 0$
$\begin{aligned} &I=2 \pi \int_{1}^{0} \frac{t d t}{t\left(1-t^{4}\right)} \\ & \end{aligned}$
$=2 \pi \int_{1}^{0} \frac{d t}{\left(1-t^{4}\right)} \\$
$=2 \pi \int_{1}^{0} \frac{d t}{(1-t)(1+t)\left(1+t^{2}\right)}$
Now by partial fraction
$\begin{aligned} &\frac{1}{(1-t)(1+t)\left(1+t^{2}\right)}=\frac{A}{(1-t)}+\frac{B}{(1+t)}+\frac{C}{\left(1+t^{2}\right)} \\ & \end{aligned}$
$1=A(1+t)\left(1+t^{2}\right)+B(1-t)\left(1+t^{2}\right)+C(1-t)(1+t)$
$\begin{aligned} &\text { Put } t=1, \text { we get } \\ & \end{aligned}$
$A=\frac{1}{4} \\$
$\text { Putting } t=-1, \text { we get }$
$\begin{aligned} &1=A+B+D \\ & \end{aligned}$
$D=1-\frac{1}{4}-\frac{1}{4} \\$
$=\frac{1}{2}$
Equating coefficients of $t^{3}$ both sides
$\begin{aligned} &A-B+C=0 \\ & \end{aligned}$
$\frac{1}{4}-\frac{1}{4}+C=0 \\$
$C=0$
$\begin{aligned} &I=2 \pi \int_{1}^{0} \frac{d t}{(1-t)(1+t)\left(1+t^{2}\right)} \\ & \end{aligned}$
$=2 \pi \int_{1}^{0} \frac{\frac{1}{4}}{1-t^{2}} d t+2 \pi \int_{1}^{0} \frac{\frac{1}{4}}{1+t} d t+2 \pi \int_{1}^{0} \frac{\frac{1}{2}}{1+t^{2}} d t$
$\begin{aligned} &=\frac{2 \pi}{4} \times\left[\frac{\log (1-t)}{-1}\right]_{1}^{0}+\frac{2 \pi}{4} \times[\log (1+t)]_{1}^{0}+\frac{2 \pi}{2} \times\left[\tan ^{-1} t\right]_{1}^{0} \\ & \end{aligned}$
$=\frac{-\pi}{2}(\log 1-\log 0)+\frac{\pi}{2}(\log 1-\log 2)+\pi\left(\tan ^{-1} 0-\tan ^{-1} 1\right)$
$\begin{aligned} &=\frac{-\pi}{2}[0-(-\infty)]+\frac{\pi}{2}[0-\log 2]+\frac{\pi}{2}\left(0-\frac{\pi}{4}\right) \\ & \end{aligned}$
$=-\infty-\frac{\pi}{2} \log 2-\frac{\pi^{2}}{4}$
$= -\infty$ [any validated up with $\infty$ becomes $\infty$ ]

Definite Integrals Exercise 19.3 question 27

Answer: $3$
Hint: You must know the rules of solving definite integral.
Given: $\int_{0}^{2} 2 x[x] d x$
Solution:
$\begin{gathered} \mathrm{I}=\int_{0}^{2} 2 x[x] d x \\ \end{gathered}$
$x=\left\{\begin{array}{l} 0 \leq x<1 \\ 1 \leq x<2 \\ 2 \leq x<3 \end{array}\right.$
$\begin{aligned} &I=\int_{0}^{1} 2 x[x]+\int_{1}^{2} 2 x[x] d x \\ & \end{aligned}$
$=\int_{1}^{2} 2 x(0) d x+\int_{1}^{2} 2 x(1) d x$
$\begin{aligned} &=\left(\frac{2 x^{2}}{2}\right)_{1}^{2} \\ & \end{aligned}$
$=4-1=3$

Definite Integrals Exercise 19.3 question 28

Answer: $\pi^{2}$
Hint: You must know about the rules of solving definite integral.
Given:
$\int_{0}^{2 \pi} \cos ^{-1}(\cos x) d x$
Solution:
$\cos ^{-1}(\cos x)=\left\{\begin{array}{c} x, 0 \leq x \leq \pi \\ 2 \pi-x, \pi \leq x \leq 2 \pi \end{array}\right\}$
$\begin{aligned} &=\int_{0}^{\pi} x d x+\int_{\pi}^{2 \pi}(2 \pi-x) d x \\ & \end{aligned}$
$=\left(\frac{x^{2}}{2}\right)_{0}^{\pi}+\left(2 \pi x-\frac{x^{2}}{2}\right)_{\pi}^{2 \pi}$
$\begin{aligned} &=\frac{\pi^{2}}{2}+4 \pi^{2}-2 \pi^{2}-2 \pi^{2}+\frac{\pi^{2}}{2} \\ & \end{aligned}$
$=\frac{\pi^{2}}{2}+\frac{\pi^{2}}{2}=\pi^{2}$

Class 12, Mathematics, chapter 19, Definite Integrals, is a part where the students take a lot of time to understand the complex concepts. The third exercise, ex 19.3, consists of 31 questions, including its subparts to be solved by the students. These questions are divided into levels 1 and 2 according to their difficulty standards. Students can find solutions for all these questions in a single RD Sharma Class 12 Chapter 19 Exercise 19.3 book whenever they have doubts.

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