Question

${\int }_0^2|1-x^2|dx$

Answer 1

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

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

$={\int }_0^{1}dx-{\int }_0^1{x}^{2}dx+{\int }_1^2{x}^{2}dx-{\int }_1^{2}dx$

$=1-\frac{1}{3}+(\frac{8}{3}-\frac{1}{3})-1=\frac{8}{3}-\frac{2}{3}$

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