## Calculus (3rd Edition)

$\frac{3}{2}$
We have $$\int_{-2}^2f(x)dx=\int_{-2}^1f(x)dx+\int_{1}^2f(x)dx=5+8=13.$$ Now, since $f(x)$ is even, then $$\int_{0}^2f(x)dx=13/2.$$ Now, $$\int_{0}^1f(x)dx=\int_{0}^2f(x)dx-\int_{1}^2f(x)dx=(13/2) -5=\frac{3}{2}.$$