Question
Evaluate the following integrals:
$\int\limits^4_{0}\big(|\text{x}|+|\text{x}+2|+|\text{x}+4|\big)\text{dx}$
$\int\limits^4_{0}\big(|\text{x}|+|\text{x}+2|+|\text{x}+4|\big)\text{dx}$
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Answer
$\text{I}=\int\limits^4_{0}\big\{|\text{x}|+|\text{x}+2|+|\text{x}+4|\big\}\text{dx}$
$\Rightarrow\text{I}=\int\limits^4_0|\text{x}|\text{dx}+\int\limits^4_0|\text{x}-2|\text{dx}+\int\limits^4_0|\text{x}-4|\text{dx}$
We know that,
$|\text{x}|=\begin{cases}-\text{x},&-5\leq\text{x}\leq0\\\text{x},&\text{x}>0\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&0\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&0\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_{0}\text{x dx}-\int\limits^2_{0}(\text{x}-2)\text{dx}+\int\limits^4_{2}(\text{x}-2)\text{dx}-\int\limits^4_{0}(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}\Big]^4_0-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^2_0+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_0$
$\Rightarrow\text{I}=8-(2-4)+8-8-2+4-(8-16)$
$\Rightarrow\text{I}=20$
$\Rightarrow\text{I}=\int\limits^4_0|\text{x}|\text{dx}+\int\limits^4_0|\text{x}-2|\text{dx}+\int\limits^4_0|\text{x}-4|\text{dx}$
We know that,
$|\text{x}|=\begin{cases}-\text{x},&-5\leq\text{x}\leq0\\\text{x},&\text{x}>0\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&0\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&0\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_{0}\text{x dx}-\int\limits^2_{0}(\text{x}-2)\text{dx}+\int\limits^4_{2}(\text{x}-2)\text{dx}-\int\limits^4_{0}(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}\Big]^4_0-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^2_0+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_0$
$\Rightarrow\text{I}=8-(2-4)+8-8-2+4-(8-16)$
$\Rightarrow\text{I}=20$
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