Let $f(x)=\left\{\begin{array}{lr}-2, & -2 \leq x \leq 0 \ x-2, & 0

Question

Let $f(x)=\left\{\begin{array}{lr}-2, & -2 \leq x \leq 0 \ x-2, & 0 < x \leq 2\end{array}ight.$ and $h(x)=f(|x|)+|f(x)|$. Then $\int_{-2}^2 h(x) \mathrm{d} x$ is equal to :

Aicharya · Accepted Answer

$h(x)=\left\{\begin{array}{cc}x-2+2-x=0, & 0 \leq x \leq 2 \ -x-2+2=-x & -2 \leq x $\Rightarrow \int_0^2 \mathrm{~h}(\mathrm{x}) \mathrm{dx}=0$ and $\int_{-2}^0 \mathrm{~h}(\mathrm{x}) \mathrm{dx}=2$

Let $f(x)=\left\{\begin{array}{lr}-2, & -2 \leq x \leq 0 \\ x-2, & 0 < x \leq 2\end{array}\right.$ and…

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