$\int_{-1}^2|x| d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} I & =\int_{-1}^2|x| d x=\int_{-1}^0(-x) d x+\int_0^2(x) d x \ & =\left[\frac{-x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}\right]_0^2=\left(0+\frac{1}{2}\right)+\left(\frac{4}{2}-0\right)=\frac{5}{2} \end{aligned}$

\(\int_{-1}^2|x| d x=\)

Solution