Question

The value of $\stackrel{2}{\underset{-2}{\int }}|1-x^2|dx$ is _____

• A. $-4$
• B. $2$
• C. $4$
• D. $-2$

Answer 1

Answer: (C)

$4$

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

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

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

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

$=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}$

$=4$

${\int }_{-2}^2|1-x^2|dx=4$.