Question

The value of the integral $$\underset{-2}{\overset{2}{\int }}|1-x^2|dx$$ is ________ .

Options

•  4

•  2

• −2

• 0

Answer 1

4

$$\text{We have},$$
$$I={\int }_{-2}^2|1-x^{{}^2}|dx$$
$$|1-x^2|=\{\begin{array}{lll}-(1-x^2)& ,& -2<x<-1\\ (1-x^2)& ,& -1<x<1\\ -(1-x^2)& ,& 1<x<2\end{array}$$
$$\therefore I={\int }_{-2}^{-1}|1-x^2|dx+{\int }_{-1}^1|1-x^2|dx+{\int }_1^2|1-x^2|dx$$
$$={\int }_{-2}^{-1}-(1-x^2)dx+{\int }_{-1}^1(1-x^2)dx+{\int }_1^2-(1-x^2)dx$$
$$=-{\int }_{-2}^{-1}(1-x^2)dx+{\int }_{-1}^1(1-x^2)dx-{\int }_1^2(1-x^2)dx$$
$$=-{[x-\frac{x^3}{3}]}_{-2}^{-1}+{[x-\frac{x^3}{3}]}_{-1}^1-{[x-\frac{x^3}{3}]}_1^2$$
$$=-[-1+\frac{1}{3}+2-\frac{8}{3}]+[1-\frac{1}{3}+1-\frac{1}{3}]-[2-\frac{8}{3}-1+\frac{1}{3}]$$
$$=-[1-\frac{7}{3}]+[2-\frac{2}{3}]-[1-\frac{7}{3}]$$
$$=-1+\frac{7}{3}+2-\frac{2}{3}-1+\frac{7}{3}$$
$$=4$$