int_0^{pi / 2} |sin t - cos t| , dt = | vedclass

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(B) Let $I = \int_0^{\pi / 2} |\sin t - \cos t| \, dt$. Since $\sin t - \cos t \le 0$ for $t \in [0, \pi / 4]$ and $\sin t - \cos t \ge 0$ for $t \in

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_0^{\pi / 2} |\sin t - \cos t| \, dt$. Since $\sin t - \cos t \le 0$ for $t \in [0, \pi / 4]$ and $\sin t - \cos t \ge 0$ for $t \in [\pi / 4, \pi / 2]$,we split the integral: $I = \int_0^{\pi / 4} (\cos t - \sin t) \, dt + \int_{\pi / 4}^{\pi / 2} (\sin t - \cos t) \, dt$. Evaluating the first part: $[\sin t + \cos t]_0^{\pi / 4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$. Evaluating the second part: $[-\cos t - \sin t]_{\pi / 4}^{\pi / 2} = (0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$. Adding both parts: $I = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$. Thus,option $B$ is correct.

$\int_0^{\pi / 2} |\sin t - \cos t| \, dt =$

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