question_answer

The value of \[\int\limits_{-2}^{3}{|1-{{x}^{2}}|dx}\] is

A) \[\frac{7}{3}\]              

B) \[\frac{14}{3}\]

C) \[\frac{28}{3}\] 

D) \[\frac{1}{3}\]

Correct Answer: C

Solution :

[c] : \[\therefore \]\[\int\limits_{-2}^{3}{|1-{{x}^{2}}|dx}\] \[=\int\limits_{-2}^{-1}{({{x}^{2}}-1)dx}+\int\limits_{-1}^{1}{-({{x}^{2}}-1)dx}+\int\limits_{1}^{3}{({{x}^{2}}-1)}dx\] \[=\left[ \left( -\frac{1}{3}+1 \right)-\left( -\frac{8}{3}+2 \right) \right]-\left[ \left( \frac{1}{3}-1 \right)-\left( -\frac{1}{3}+1 \right) \right]\] \[+\left[ (9-3)-\left( \frac{1}{3}-1 \right) \right]\] \[=\frac{4}{3}+\frac{4}{3}+6+\frac{2}{3}=\frac{28}{3}\]