RD Sharma Class 12 Exercise 19.3 Definite Integrals Solutions Maths

Q: Why is RD Sharma Books used by most of the Class 12 students who are preparing for the public examinations?

Firstly, the solutions provided in these books are given by the experts, verified to be accurate. Secondly, there were instances when practicing questions from the RD Sharma Class 12th Exercise 19.3 portion being asked for the public exam. Hence, many students use these books.

Q: Is it possible to find the RD Sharma solution books for free of cost?

Visit the Career 360 website to access the best set of RD Sharma solution books for free of cost. You need not make any payment to access these books.

Q: Can the RD Sharma solution books be downloaded?

The RD Sharma Class 12 Solutions Definite Integrals Ex 19.3 book can be download to your device. You need to hit the Download button present on the screen near the book you require.

Q: How can I view the RD Sharma solution book for the class 12 chapter 19, ex 19.3 portions?

Visit the Career 360 website and type the name of the book in the Search box. The results will be displayed accordingly. In this case, type RD Sharma Class 12th Exercise 19.3 solution book to find the answer key accordingly.

Q: Can CBSE board students refer to the RD Sharma solution books to clear their doubts?

The RD Sharma solution books follow the NCERT pattern; hence the CBSE board students can very well refer to these books to clarify their doubts.

RD Sharma Class 12 Exercise 19.3 Definite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:22 PM IST

The class 12 students require a proper guide other than their teachers and tutors to help them clarify their doubts. The RD Sharma books are one of the best solution materials recommended by many CBSE schools to their students. However, most of the Class 12 students find it challenging to solve the mathematics, chapter 19 sums. In such cases, the RD Sharma Class 12th Exercise 19.3 is essential.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter19 Definite Integrals - Other Exercise

Chapter 19 - Definite Integrals - Ex-19.1

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

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.

A group of staff members who are experts in their respective domains has created solutions for these questions. They have provided solutions in every way possible for a single sum. As it follows the NCERT patterns, CBSE school students can use the RD Sharma books to clarify their doubts. This RD Sharma Class 12th Exercise 19.13 solution book also consists of various practice questions that help the students work out and gain confidence to face the challenging questions in the exams.

Definite Integrals would no more be a complex concept once practiced well. The students can use this resource material, Class 12 RD Sharma Chapter 19 Exercise 19.3 Solution, to do their homework and assignment. They can also use it to prepare for their tests and examinations. With a copy of RD Sharma Class 12th Exercise 19.3 solution book, the students need not have a teacher or tutor by their side always. RD Sharma's book offers excellent guidance in solving students' doubts.

The most significant advantage of using RD Sharma books is that they can be accessed for free of cost at the Career360 website. The option to download the RD Sharma Class 12th Exercise 19.3 material is also available. Lots of students have already benefitted by scoring high marks using RD Sharma books for their exam preparation.

There are chances that questions for the public exams will be picked from the RD Sharma books. Hence, for the students preparing for their exams, using the RD Sharma Class 12 Solutions Chapter 19 Ex 19.3 book from day one will help them score better.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. Why is RD Sharma Books used by most of the Class 12 students who are preparing for the public examinations?

Firstly, the solutions provided in these books are given by the experts, verified to be accurate. Secondly, there were instances when practicing questions from the RD Sharma Class 12th Exercise 19.3 portion being asked for the public exam. Hence, many students use these books.

2. Is it possible to find the RD Sharma solution books for free of cost?

Visit the Career 360 website to access the best set of RD Sharma solution books for free of cost. You need not make any payment to access these books.

3. Can the RD Sharma solution books be downloaded?

The RD Sharma Class 12 Solutions Definite Integrals Ex 19.3 book can be download to your device. You need to hit the Download button present on the screen near the book you require.

4. How can I view the RD Sharma solution book for the class 12 chapter 19, ex 19.3 portions?

Visit the Career 360 website and type the name of the book in the Search box. The results will be displayed accordingly. In this case, type RD Sharma Class 12th Exercise 19.3 solution book to find the answer key accordingly.

5. Can CBSE board students refer to the RD Sharma solution books to clear their doubts?

The RD Sharma solution books follow the NCERT pattern; hence the CBSE board students can very well refer to these books to clarify their doubts.