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

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

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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

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