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#### Need solution for RD Sharma Maths Class 12 Chapter 19 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

\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}

$\Rightarrow I=8-(2-4)+8-8-2+4-(8-16) \\$

$\Rightarrow I=20$