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

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

$|x-2|=\left\{\begin{array}{l} -(x-2), 1 \leq x \leq 2 \\ x-2,2 \leq x \leq 4 \end{array}\right\} \\$

$|x-3|=\left\{\begin{array}{l} -(x-4), 1 \leq x \leq 4 \\ x-4, x>4 \end{array}\right\}$

$\int_{1}^{4}(x-1)+\int_{1}^{2}-(x-2) d x+\int_{2}^{4}(x-2) d x+\int_{1}^{4}-(x-4) d x$

\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}

$=\left(\frac{16}{2}-4-\frac{1}{2}+1\right)+\left(-2+4+\frac{1}{2}-2\right)+\left(\frac{16}{2}-8-2+4\right)+\left(-\frac{16}{2}+16+\frac{1}{2}-4\right) \\$

$=\frac{23}{2}$