The value of int_{-2}^{3} |1 - x^2| dx is | vedclass

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(D) We need to evaluate the integral $I = \int_{-2}^{3} |1 - x^2| dx$.The expression $|1 - x^2|$ changes sign at $x = -1$ and $x = 1$.For $x \in [-2,

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$\frac{28}{3}$

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(D) We need to evaluate the integral $I = \int_{-2}^{3} |1 - x^2| dx$.The expression $|1 - x^2|$ changes sign at $x = -1$ and $x = 1$.For $x \in [-2, -1]$,$1 - x^2 \le 0$,so $|1 - x^2| = x^2 - 1$.For $x \in [-1, 1]$,$1 - x^2 \ge 0$,so $|1 - x^2| = 1 - x^2$.For $x \in [1, 3]$,$1 - x^2 \le 0$,so $|1 - x^2| = x^2 - 1$.Thus,$I = \int_{-2}^{-1} (x^2 - 1) dx + \int_{-1}^{1} (1 - x^2) dx + \int_{1}^{3} (x^2 - 1) dx$.Evaluating each integral:$\int_{-2}^{-1} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{-2}^{-1} = (-\frac{1}{3} + 1) - (-\frac{8}{3} + 2) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.$\int_{-1}^{1} (1 - x^2) dx = [x - \frac{x^3}{3}]_{-1}^{1} = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.$\int_{1}^{3} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{1}^{3} = (9 - 3) - (\frac{1}{3} - 1) = 6 - (-\frac{2}{3}) = \frac{20}{3}$.Summing these values: $I = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}$.

The value of $\int_{-2}^{3} |1 - x^2| dx$ is

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