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

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(A) We need to evaluate the integral $I = \int_{-2}^3 |1-x^2| dx$. The expression inside the absolute value,$1-x^2$,changes sign at $x = -1$ and $x =

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$\frac{28}{3}$

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(A) We need to evaluate the integral $I = \int_{-2}^3 |1-x^2| dx$. The expression inside the absolute value,$1-x^2$,changes sign at $x = -1$ and $x = 1$. We split the integral into three intervals: $[-2, -1]$,$[-1, 1]$,and $[1, 3]$. In $[-2, -1]$,$1-x^2 \le 0$,so $|1-x^2| = x^2-1$. In $[-1, 1]$,$1-x^2 \ge 0$,so $|1-x^2| = 1-x^2$. 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 part: $\int_{-2}^{-1} (x^2-1) dx = [\frac{x^3}{3} - x]_{-2}^{-1} = (-\frac{1}{3} + 1) - (-\frac{8}{3} + 2) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$. $\int_{-1}^1 (1-x^2) dx = [x - \frac{x^3}{3}]_{-1}^1 = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$. $\int_{1}^3 (x^2-1) dx = [\frac{x^3}{3} - x]_1^3 = (9 - 3) - (\frac{1}{3} - 1) = 6 - (-\frac{2}{3}) = \frac{20}{3}$. Summing these values: $I = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}$.

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

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