The area of the region under the curve $y=|\sin x-\cos x|$, $0 \leq x \leq \frac{\pi}{2}$ and above $x$-axis, is (in square units)

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } y=|\sin x-\cos x|, 0 \leq x \leq \frac{\pi}{2} \ & 	ext { Now, Area }=\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x \ & =\int_0^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x \ & =[\sin x+\cos x]_0^{\frac{\pi}{4}}-[\cos x+\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ & =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1-\left[1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}ight]=2(\sqrt{2}-1)\end{aligned}$

The area of the region under the curve $y=|\sin x-\cos x|$, $0 \leq x \leq \frac{\pi}{2}$ and above $x$-axis…

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