A) \[0\]
B) \[2(\sqrt{2}-1)\]
C) \[\sqrt{2}-1\]
D) \[2(\sqrt{2}+1)\]
Correct Answer: B
Solution :
\[\int_{0}^{\pi /2}{|\sin x-\cos x|dx}\] \[=\int_{0}^{\pi /4}{-(\sin x-\cos x)dx}\] \[+\int_{\pi /4}^{\pi /2}{-(\sin x-\cos x)dx}\] \[=-[-\cos x-\sin x]_{0}^{\pi /4}\] \[+[-\cos x-\sin x]_{\pi /4}^{\pi /2}\] \[=-\left[ -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1 \right]\] \[+\left[ -0-1+\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right) \right]\] \[=\sqrt{2}-1-1+\sqrt{2}\] \[=2(\sqrt{2}-1)\]
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