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