A) \[\frac{28}{3}\]
B) \[\frac{14}{3}\]
C) \[\frac{7}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
\[\int_{-2}^{3}{|1-{{x}^{2}}|}dx=\int_{-2}^{-1}{({{x}^{2}}-1)}\,dx\] \[+\int_{-1}^{1}{(1-{{x}^{2}})dx}+\int_{1}^{3}{({{x}^{2}}-1)}dx\] \[=\left[ \frac{{{x}^{3}}}{3}-x \right]_{-2}^{-1}+\left[ x-\frac{{{x}^{3}}}{3} \right]_{-1}^{1}+\left[ \frac{{{x}^{3}}}{3}-x \right]_{1}^{3}\] \[=\left[ -\frac{1}{3}+1+\frac{8}{3}-2 \right]+\left[ 1-\frac{1}{3}+1-\frac{1}{3} \right]\] \[+\left[ 9-3-\frac{1}{3}+1 \right]\] \[=\frac{4}{3}+\frac{4}{3}+\frac{20}{3}\] \[=\frac{28}{3}\]
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