Question

The value of ${\int }_0^2\left(x\right\}dx$, {x} denotes fractional part of x is

Answer 1

${\int }_0^2\left(x\right\}dx={\int }_0^{1}xdx+{\int }_1^2\left(x-1\right)dx={\left(\frac{x^2}{2}\right]}_0^1+{\left(\frac{{\left(x-1\right)}^2}{2}\right]}_1^2=\frac{1}{2}+\frac{1}{2}=1$