Question

Evaluate ${\int }_0^2\left\{x\right\}dx$, (where $\left\{x\right\}$ denotes the fractional part of $x$.)

Answer 1

We have,
${\int }_0^2\left\{x\right\}dx={\int }_0^2(x-[x\left]\right)dx={\int }_0^{2}xdx-{\int }_0^2\left[x\right]dx$
$={\left(\frac{x^2}{2}\right)}_0^2-({\int }_0^1\left[x\right]dx+{\int }_1^2\left[x\right]dx)$
$=\frac{1}{2}(4-0)-({\int }_0^{1}0dx+{\int }_1^{2}1dx)$
$=2-{\left(x\right)}_1^2=2-\left(1\right)=1$