The value of $\int_{-2}^3\left|1-x^2\right| d x$ is
The value of $\int_{-2}^3\left|1-x^2\right| d x$ is
$\frac{28}{3}$
$\frac{14}{3}$
$\frac{7}{3}$
$\frac{1}{3}$
Solution
$\int_{-2}^{-1}\left(x^2-1\right) d x+\int_{-1}^1\left(1-x^2\right) d x+\int_1^3\left(x^2-1\right) d x=\frac{x^3}{3}-\left.x\right|_{-2} ^{-1}+x-\left.\frac{x^3}{3}\right|_{-1} ^1+\frac{x^3}{3}-\left.x\right|_1 ^3=\frac{28}{3}$