int_0^2 |1-x^2| dx = | vedclass

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(B) We need to evaluate the definite integral $\int_0^2 |1-x^2| dx$.First, we analyze the absolute value function $|1-x^2|$. The expression $1-x^2$ ch

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$2$

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(B) We need to evaluate the definite integral $\int_0^2 |1-x^2| dx$.First, we analyze the absolute value function $|1-x^2|$. The expression $1-x^2$ changes sign at $x=1$ within the interval $[0, 2]$.For $0 \leq x For $1 \leq x \leq 2$, $1-x^2 \leq 0$, so $|1-x^2| = -(1-x^2) = x^2-1$.Using the property of definite integrals, we split the integral at $x=1$:$\int_0^2 |1-x^2| dx = \int_0^1 (1-x^2) dx + \int_1^2 (x^2-1) dx$Now, we integrate each part:$\int_0^1 (1-x^2) dx = [x - \frac{x^3}{3}]_0^1 = (1 - \frac{1}{3}) - 0 = \frac{2}{3}$$\int_1^2 (x^2-1) dx = [\frac{x^3}{3} - x]_1^2 = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} - (-\frac{2}{3}) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$Adding these results together:$\frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2$

$\int_0^2 |1-x^2| dx = $

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