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Question

(A) The shaded region is bounded by $y = \sin x$ from $x = 2\pi$ to $x = \frac{9\pi}{4}$ and by $y = \cos x$ from $x = \frac{9\pi}{4}$ to $x = \frac{5

Vedclass · Accepted Answer

$2 - \sqrt{2}$

Vedclass · Answer

(A) The shaded region is bounded by $y = \sin x$ from $x = 2\pi$ to $x = \frac{9\pi}{4}$ and by $y = \cos x$ from $x = \frac{9\pi}{4}$ to $x = \frac{5\pi}{2}$.The area $A$ is given by:$A = \int_{2\pi}^{\frac{9\pi}{4}} \sin x \, dx + \int_{\frac{9\pi}{4}}^{\frac{5\pi}{2}} \cos x \, dx$$A = [-\cos x]_{2\pi}^{\frac{9\pi}{4}} + [\sin x]_{\frac{9\pi}{4}}^{\frac{5\pi}{2}}$$A = -(\cos \frac{9\pi}{4} - \cos 2\pi) + (\sin \frac{5\pi}{2} - \sin \frac{9\pi}{4})$$A = -(\frac{1}{\sqrt{2}} - 1) + (1 - \frac{1}{\sqrt{2}})$$A = 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}}$$A = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$ sq. units.

The area of the shaded region is ... sq. units.

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