Question
Find the value of $\int_{-4}^4|x| d x$.

Answer

Let $ I =\int_{-4}^4|x| d x=\int_{-4}^0|x| d x+\int_0^4|x| d x$
$ \because|x|=\left\{\begin{array}{l} x, x>0 \\ -x, x<0\end{array}\right.$
$\therefore =\int_{-4}^0 - x\ d x+\int_0^4 x\ d x$
$ =\left(\frac{-x^2}{2}\right)_{-4}^0+\left(\frac{x^2}{2}\right)_0^4$
$ =\frac{-1}{2}(0-16)+\frac{1}{2}(16-0)$
$=8+8$
$=16$

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