The value of $\int\limits_{-2}^3\left|1-x^2\right| d x$, is

$\frac{28}{3}$

We have,

$\left|1-x^2\right|=\left|x^2-1\right|=\left\{\begin{array}{r}-\left(x^2-1\right) \text { if }-1<x<1 \\ x^2-1, \text { otherwise }\end{array}\right.$

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

$\Rightarrow I=\int_{-2}^{-1}\left(x^2-1\right) d x+\int_{-1}^1-\left(x^2-1\right) d x+\int_1^3\left(x^2-1\right) d x$

$\Rightarrow I=\left[\frac{x^3}{3}-x\right]_{-2}^{-1}-\left[\frac{x^3}{3}-x\right]_{-1}^1+\left[\frac{x^3}{3}-x\right]_1^3$

$\Rightarrow I=\left[\left(-\frac{1}{3}+1\right)-\left(-\frac{8}{3}+2\right)\right]-\left[\left(\frac{1}{3}-1\right)-\left(-\frac{1}{3}+1\right)\right] +\left[\left(\frac{27}{3}-3\right)-\left(\frac{1}{3}-1\right)\right]$

$\Rightarrow I=\frac{4}{3}+\frac{4}{3}+\frac{20}{3}=\frac{28}{3}$