Question

Solve ${\int }_0^2(\left|x\right|+|x-1|)dx=$

Answer 1

Let

$I={\int }_0^{2}xdx-{\int }_0^1\left(x-1\right)+{\int }_0^1\left(x-1\right)$

$={\left[\frac{x^2}{2}\right]}_0^2-{\left[\frac{x^2}{2}-x\right]}_0^1+{\left[\frac{x^2}{2}-x\right]}_1^2$

$=2+\frac{1}{2}+\frac{1}{2}=3$