∴I=∫−23 1−x2dx=∫−2−1 x2−1dx+∫−11 −x2−1dx+∫13 x2−1dx=x33−x−2−1−x33−x−11+x33−x13=−13+1−−83+2−13−1−−13+1+273−3−13−1=43+43+203=283

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$\int_{- 2}^{3} |1 - x^{2}| dx =$

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$\frac{7}{3}$

$\frac{14}{3}$

$\frac{28}{3}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} \begin{array}{l} ∴ I = \int_{- 2}^{3} |1 - x^{2}| d x \\ = \int_{- 2}^{- 1} (x^{2} - 1) d x + \int_{- 1}^{1} - (x^{2} - 1) d x + \int_{1}^{3} (x^{2} - 1) d x \\ = {[\frac{x^{3}}{3} - x]}_{- 2}^{- 1} - {[\frac{x^{3}}{3} - x]}_{- 1}^{1} + {[\frac{x^{3}}{3} - x]}_{1}^{3} \\ = [(- \frac{1}{3} + 1) - (- \frac{8}{3} + 2)] - [(\frac{1}{3} - 1) - (- \frac{1}{3} + 1)] + [(\frac{27}{3} - 3) - (\frac{1}{3} - 1)] \\ = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3} \end{array} \end{array}$