Evaluate the following integrals: ^4_1 |x-1|+|x-2|+|x-4 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^4_1\big\{|\text{x}-1|+|\text{x}-2|+|\text{x}-4\big\}\text{dx}$

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Answer

$\text{I}=\int\limits^4_1\big\{|\text{x}-1|+|\text{x}-2|+|\text{x}-4\big\}\text{dx}$
$\Rightarrow\text{I}=\int\limits^4_1|\text{x}-1|\text{dx}+\int\limits^4_1|\text{x}-2|\text{dx}+\int\limits^4_1|\text{x}-4|\text{dx}$
We know that,
$|\text{x}-2|=\begin{cases}-(\text{x}-1),&\text{x}\leq1\\\text{x}-1,&1<\text{x}\leq4\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&1\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&1\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_1(\text{x}-1)\text{dx}-\int\limits^2_1(\text{x}-2)\text{dx}+\int\limits^4_2(\text{x}-2)\text{dx}-\int\limits^4_1(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_1-\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^2_1+\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_1$
$\Rightarrow\text{I}=8-4-\frac{1}{2}+1-\Big(2-4-\frac{1}{2}+2\Big)+8-8-2+4-\Big(8-16-\frac{1}{2}+4\Big)$
$\Rightarrow\text{I}=\frac{23}{2}$

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