The area of the shaded region bounded by the | vedclass

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(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

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$2 - \sqrt{2}$

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 intersection point is where $\sin x = \cos x$, which is $	an x = 1$, so $x = \frac{9\pi}{4}$.The area $A$ is given by:$A = \int_{2\pi}^{9\pi/4} \sin x \, dx + \int_{9\pi/4}^{5\pi/2} \cos x \, dx$$A = [-\cos x]_{2\pi}^{9\pi/4} + [\sin x]_{9\pi/4}^{5\pi/2}$$A = -(\cos(9\pi/4) - \cos(2\pi)) + (\sin(5\pi/2) - \sin(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}} = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$ sq. units.

The area of the shaded region bounded by the curves $y = \sin x$, $y = \cos x$ and the $x$-axis between $x = 2\pi$ and $x = \frac{5\pi}{2}$ is ... sq. units.

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