A) 4
B) 2
C) \[-2\]
D) 0
E) 1
Correct Answer: A
Solution :
\[\int_{-2}^{2}{|1-x{{|}^{2}}}dx=2\int_{0}^{2}{|1-{{x}^{2}}|}\,dx\] \[=2[\int_{0}^{1}{{{(1-x)}^{2}}x+\int_{1}^{2}{({{x}^{2}}-1)}dx}]\] \[=2\left[ \left( x-\frac{{{x}^{3}}}{3} \right)_{0}^{1}+\left( \frac{{{x}^{3}}}{3}-x \right)_{1}^{2} \right]\] \[=2\left[ \frac{2}{3}+\frac{2}{3}+\frac{2}{3} \right]=4\]You need to login to perform this action.
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