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  • Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 17 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 17 Maths Textbook Solution.

Answers (1)

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}

 

Posted by

infoexpert27

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