Browse by Stream
QnA
Home

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 20

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 20

Answers (1)

best_answer

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}

        

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board 2025: Class 10th, 12th supplementary date sheet out; exams from July 15
anam-khan

CBSE issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee
anam-khan

CBSE Supplementary Exam 2025: Registrations open for Class 10, 12 private students today

Latest Question