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

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(B) The function $f(x) = |1 - x^2|$ changes its sign at $x = -1$ and $x = 1$. We can split the integral as: $\int_{-2}^{2} |1 - x^2| \, dx = \int_{-2}

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The function $f(x) = |1 - x^2|$ changes its sign at $x = -1$ and $x = 1$. We can split the integral as: $\int_{-2}^{2} |1 - x^2| \, dx = \int_{-2}^{-1} |1 - x^2| \, dx + \int_{-1}^{1} |1 - x^2| \, dx + \int_{1}^{2} |1 - x^2| \, dx$ In the interval $[-2, -1]$,$1 - x^2 \le 0$,so $|1 - x^2| = -(1 - x^2) = x^2 - 1$. In the interval $[-1, 1]$,$1 - x^2 \ge 0$,so $|1 - x^2| = 1 - x^2$. In the interval $[1, 2]$,$1 - x^2 \le 0$,so $|1 - x^2| = -(1 - x^2) = x^2 - 1$. Thus,the integral becomes: $\int_{-2}^{-1} (x^2 - 1) \, dx + \int_{-1}^{1} (1 - x^2) \, dx + \int_{1}^{2} (x^2 - 1) \, dx$ $= [\frac{x^3}{3} - x]_{-2}^{-1} + [x - \frac{x^3}{3}]_{-1}^{1} + [\frac{x^3}{3} - x]_{1}^{2}$ $= ((\frac{-1}{3} + 1) - (\frac{-8}{3} + 2)) + ((1 - \frac{1}{3}) - (-1 + \frac{1}{3})) + ((\frac{8}{3} - 2) - (\frac{1}{3} - 1))$ $= (\frac{2}{3} - (-\frac{2}{3})) + (\frac{2}{3} - (-\frac{2}{3})) + (\frac{2}{3} - (-\frac{2}{3}))$ $= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4.$

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

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