Question

The value of the integral $\underset{-2}{\overset{2}{\int }}\left|1-x^2\right|dx$ is
(a) 4
(b) 2
(c) −2
(d) 0

Answer 1

(a) 4

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_{-2}^2\left|1-x^{{}^2}\right|dx \\ \left|1-x^{{}^2}\right|=\left\{\begin{array}{l}\begin{array}{l}-\left(1-x^{{}^2}\right),-2<x<-1\\ \left(1-x^{{}^2}\right),-1<x<1\end{array}\\ -\left(1-x^{{}^2}\right),1<x<2\end{array}\right. \\ \therefore I={\int }_{-2}^{-1}\left|1-x^{{}^2}\right|dx+{\int }_{-1}^1\left|1-x^{{}^2}\right|dx+{\int }_1^2\left|1-x^{{}^2}\right|dx \\ ={\int }_{-2}^{-1}-\left(1-x^{{}^2}\right)dx+{\int }_{-1}^1\left(1-x^{{}^2}\right)dx+{\int }_1^2-\left(1-x^{{}^2}\right)dx \\ =-{\int }_{-2}^{-1}\left(1-x^{{}^2}\right)dx+{\int }_{-1}^1\left(1-x^{{}^2}\right)dx-{\int }_1^2\left(1-x^{{}^2}\right)dx \\ =-{\left[x-\frac{x^3}{3}\right]}_{-2}^{-1}+{\left[x-\frac{x^3}{3}\right]}_{-1}^1-{\left[x-\frac{x^3}{3}\right]}_1^2 \\ =-\left[-1+\frac{1}{3}+2-\frac{8}{3}\right]+\left[1-\frac{1}{3}+1-\frac{1}{3}\right]-\left[2-\frac{8}{3}-1+\frac{1}{3}\right] \\ =-\left[1-\frac{7}{3}\right]+\left[2-\frac{2}{3}\right]-\left[1-\frac{7}{3}\right] \\ =-1+\frac{7}{3}+2-\frac{2}{3}-1+\frac{7}{3} \\ =4 \end{gathered}$$