The correct evaluation of int_0^{pi /2} {left | vedclass

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(B) Let $I = \int_0^{\pi /2} {\left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right|} \,dx$. Since $\sin(x - \pi/4) \le 0$ for $x \in [0, \pi/4]$ a

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$2 - \sqrt{2}$

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(B) Let $I = \int_0^{\pi /2} {\left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right|} \,dx$. Since $\sin(x - \pi/4) \le 0$ for $x \in [0, \pi/4]$ and $\sin(x - \pi/4) \ge 0$ for $x \in [\pi/4, \pi/2]$,we split the integral: $I = - \int_0^{\pi /4} {\sin \left( {x - \frac{\pi }{4}} \right)dx} + \int_{\pi /4}^{\pi /2} {\sin \left( {x - \frac{\pi }{4}} \right)dx}$. Evaluating the integrals: $I = - [-\cos(x - \pi/4)]_0^{\pi/4} + [-\cos(x - \pi/4)]_{\pi/4}^{\pi/2}$. $I = [\cos(x - \pi/4)]_0^{\pi/4} - [\cos(x - \pi/4)]_{\pi/4}^{\pi/2}$. $I = (\cos(0) - \cos(-\pi/4)) - (\cos(\pi/4) - \cos(0))$. $I = (1 - 1/\sqrt{2}) - (1/\sqrt{2} - 1)$. $I = 1 - 1/\sqrt{2} - 1/\sqrt{2} + 1 = 2 - 2/\sqrt{2} = 2 - \sqrt{2}$.

The correct evaluation of $\int_0^{\pi /2} {\left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right|} \,dx$ is

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