int_{0}^{3} {|2 - x| , dx} equals | vedclass

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(B) Let $I = \int_{0}^{3} {|2 - x| \, dx}$.Since $|2 - x| = 2 - x$ for $x \le 2$ and $|2 - x| = -(2 - x) = x - 2$ for $x > 2$,we split the integral at

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$5/2$

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(B) Let $I = \int_{0}^{3} {|2 - x| \, dx}$.Since $|2 - x| = 2 - x$ for $x \le 2$ and $|2 - x| = -(2 - x) = x - 2$ for $x > 2$,we split the integral at $x = 2$:$I = \int_{0}^{2} {(2 - x) \, dx} + \int_{2}^{3} {(x - 2) \, dx}$Evaluating the first integral:$\int_{0}^{2} {(2 - x) \, dx} = [2x - \frac{x^2}{2}]_{0}^{2} = (4 - 2) - (0 - 0) = 2$Evaluating the second integral:$\int_{2}^{3} {(x - 2) \, dx} = [\frac{x^2}{2} - 2x]_{2}^{3} = (\frac{9}{2} - 6) - (\frac{4}{2} - 4) = (\frac{9-12}{2}) - (2 - 4) = -\frac{3}{2} - (-2) = -\frac{3}{2} + 2 = \frac{1}{2}$Adding the two parts:$I = 2 + \frac{1}{2} = \frac{5}{2}$.

$\int_{0}^{3} {|2 - x| \, dx}$ equals

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