$\left\{x\right\}=\left\{\begin{array}{l}x 0\le x<1\\ \\ x-1 1\le x<2\end{array}\right.$

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

Since ${\int }_a^{b}f\left(x\right)dx = {\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx$ where $a<c<b$

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

$={\left[x-\frac{x^3}{3}\right]}_0^1+{\left[\frac{x^3}{3}-x\right]}_1^2$

$=\left[1-\frac{1}{3}\right]-0+\left[\frac{8}{3}-2\right]-\left[\frac{1}{3}-1\right]$

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

$=\frac{6}{3}=2$