A) \[4\]
B) \[2\]
C) \[-2\]
D) \[0\]
Correct Answer: A
Solution :
Let\[I=\int_{-2}^{2}{|1-{{x}^{2}}|dx}\] \[=\int_{-2}^{-1}{({{x}^{2}}-1)dx+\int_{-1}^{1}{(1-{{x}^{2}})dx}}\] \[+\int_{1}^{2}{({{x}^{2}}-1)dx}\] \[=\left[ \frac{{{x}^{3}}}{3}-x \right]_{-2}^{-1}+\left[ x-\frac{{{x}^{3}}}{3} \right]_{-1}^{1}+\left[ \frac{{{x}^{3}}}{3}-x \right]_{1}^{2}\] \[=\left[ -\frac{1}{3}+1-\left( -\frac{8}{3}+2 \right) \right]\] \[+\left[ 1-\frac{1}{3}-\left( -1+\frac{1}{3} \right) \right]+\left[ \frac{8}{3}-2-\left( \frac{1}{3}-1 \right) \right]\] \[=\left( \frac{7}{3}-1 \right)+\left( 2-\frac{2}{3} \right)+\left( \frac{7}{3}-1 \right)\] \[=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4\]
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