Question
The value of $$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} $$ is-
A.
$$\frac{1}{3}$$
B.
$$\frac{14}{3}$$
C.
$$\frac{7}{3}$$
D.
$$\frac{28}{3}$$
Answer :
$$\frac{28}{3}$$
Solution :
$$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|dx} $$
\[{\rm{Now }}\left| {{x^2} - 1} \right| = \left\{ \begin{array}{l}
{x^2} - 1\,\,\,{\rm{if}}\,\,\,x \le - 1\\
1 - {x^2}\,\,\,{\rm{if}}\,\,\, - 1 \le x \le 1\\
{x^2} - 1\,\,\,{\rm{if}}\,\,\,x \ge 1
\end{array} \right.\]
$$\eqalign{
& \therefore \,\,\,{\text{Integral is}} \cr
& \int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx} + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)dx} + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx} \cr
& = \left[ {\frac{{{x^3}}}{3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1 + \left[ {\frac{{{x^3}}}{3} - x} \right]_1^3 \cr
& = \left( { - \frac{1}{3} + 1} \right) - \left( { - \frac{8}{3} + 2} \right) + \left( {2 - \frac{2}{3}} \right) + \left( {\frac{{27}}{3} - 3} \right) - \left( {\frac{1}{3} - 1} \right) \cr
& = \frac{2}{3} + \frac{2}{3} + \frac{4}{3} + 6 + \frac{2}{3} \cr
& = \frac{{28}}{3} \cr} $$