A) \[\frac{1}{3}\]
B) \[\frac{14}{3}\]
C) \[\frac{7}{3}\]
D) \[\frac{28}{3}\]
Correct Answer: D
Solution :
\[\int_{-2}^{3}{|1-{{x}^{2}}|dx=\int_{-2}^{-1}{({{x}^{2}}-1)dx+\int_{-1}^{1}{(1-{{x}^{2}})dx+\int_{1}^{3}{({{x}^{2}}-1)dx}}}}\] =\[\left[ \frac{{{x}^{2}}}{3}-x \right]_{-2}^{-1}+\left[ x-\frac{{{x}^{2}}}{3} \right]_{-1}^{1}+\left[ \frac{{{x}^{2}}}{3}-x \right]_{1}^{2}\] \[=\frac{2}{3}+\frac{2}{3}+2\left( \frac{2}{3} \right)+(9-3)-\left( \frac{1}{3}-1 \right)\]\[=\frac{10}{3}+6=\frac{28}{3}\].
You need to login to perform this action.
You will be redirected in
3 sec
