int_0^{pi /2} |sin x - cos x| , dx = | vedclass

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(B) We need to evaluate the integral $I = \int_0^{\pi /2} |\sin x - \cos x| \, dx$. Since $\sin x - \cos x \le 0$ for $x \in [0, \pi/4]$ and $\sin x -

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$2(\sqrt{2} - 1)$

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(B) We need to evaluate the integral $I = \int_0^{\pi /2} |\sin x - \cos x| \, dx$. Since $\sin x - \cos x \le 0$ for $x \in [0, \pi/4]$ and $\sin x - \cos x \ge 0$ for $x \in [\pi/4, \pi/2]$,we split the integral at $x = \pi/4$: $I = \int_0^{\pi /4} -(\sin x - \cos x) \, dx + \int_{\pi /4}^{\pi /2} (\sin x - \cos x) \, dx$ $I = \int_0^{\pi /4} (\cos x - \sin x) \, dx + \int_{\pi /4}^{\pi /2} (\sin x - \cos x) \, dx$ Evaluating the first part: $[\sin x + \cos x]_0^{\pi /4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$. Evaluating the second part: $[-\cos x - \sin x]_{\pi /4}^{\pi /2} = (0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$. Adding both parts: $I = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$.

$\int_0^{\pi /2} |\sin x - \cos x| \, dx = $

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