∫−23 x2−1dx is equal to

# ${\int }_{-2}^{3} \left|{\mathrm{x}}^{2}-1\right|\mathrm{dx}$ is equal to

1. A

3

2. B

$\frac{1}{3}$

3. C

$\frac{17}{3}$

4. D

$\frac{28}{3}$

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### Solution:

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

[here, modulus function will change at the points, ,when

so,

$\begin{array}{l}\mathrm{I}={\int }_{-2}^{-1} \left({\mathrm{x}}^{2}-1\right)\mathrm{dx}+{\int }_{-1}^{1} \left(1-{\mathrm{x}}^{2}\right)\mathrm{dx}+{\int }_{1}^{3} \left({\mathrm{x}}^{2}-1\right)\mathrm{dx}\\ ={\left[\frac{{\mathrm{x}}^{3}}{3}-\mathrm{x}\right]}_{-2}^{-1}+{\left[\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3}\right]}_{-1}^{1}+{\left[\frac{{\mathrm{x}}^{3}}{3}-\mathrm{x}\right]}_{1}^{3}\\ =\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+6+\frac{2}{3}=\frac{28}{3}\end{array}$

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