$$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx$$

Now $$\left| {{x^2} - 1} \right| = \left\{ {\matrix{ {{x^2} - 1} & {if} & {x \le - 1} \cr {1 - {x^2}} & {if} & { - 1 \le x \le 1} \cr {{x^2} - 1} & {if} & {x \ge 1} \cr } } \right.$$

$$\therefore$$ Integral is

$$\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)} } dx + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx} $$

$$\left[ {{{{x^3}} \over 3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - {{{x^3}} \over 3}} \right]_{ - 1}^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^3$$

$$ = \left( { - {1 \over 3} + 1} \right) - \left( { - {8 \over 3} + 2} \right) + \left( {2 - {2 \over 3}} \right) + \left( {{{27} \over 3} - 3} \right) - \left( {{1 \over 3} - 1} \right)$$

$$ = {2 \over 3} + {2 \over 3} + {4 \over 3} + 6 + {2 \over 3} = {{28} \over 3}$$