The area under the curve y = |cos x - sin x|, | vedclass

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Question

(B) The function is $y = |\cos x - \sin x|$.In the interval $[0, \pi/4]$,$\cos x \geq \sin x$,so $y = \cos x - \sin x$.In the interval $[\pi/4, \pi/2]

Vedclass · Accepted Answer

$2\sqrt{2} - 2$

Vedclass · Answer

(B) The function is $y = |\cos x - \sin x|$.In the interval $[0, \pi/4]$,$\cos x \geq \sin x$,so $y = \cos x - \sin x$.In the interval $[\pi/4, \pi/2]$,$\sin x \geq \cos x$,so $y = \sin x - \cos x$.The required area is given by:$A = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$$A = [\sin x + \cos x]_{0}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$$A = ((\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0))) + ((-\cos(\pi/2) - \sin(\pi/2)) - (-\cos(\pi/4) - \sin(\pi/4)))$$A = ((\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)) + ((0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}))$$A = (\sqrt{2} - 1) + (-1 + \sqrt{2})$$A = 2\sqrt{2} - 2$ square units.

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

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