The area enclosed by the curves y = cos x,y = | vedclass

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(C) The area $A$ is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = \frac{3\pi}{2}$.$A = \int_{0}^{\frac{3\pi}{2}}

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$2 + \frac{3\pi}{2}$

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(C) The area $A$ is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = \frac{3\pi}{2}$.$A = \int_{0}^{\frac{3\pi}{2}} ((1 + \sin 2x) - \cos x) \, dx$$= \int_{0}^{\frac{3\pi}{2}} (1 + \sin 2x - \cos x) \, dx$$= \left[ x - \frac{1}{2} \cos 2x - \sin x \right]_{0}^{\frac{3\pi}{2}}$$= \left( \frac{3\pi}{2} - \frac{1}{2} \cos(3\pi) - \sin\left(\frac{3\pi}{2}\right) \right) - \left( 0 - \frac{1}{2} \cos(0) - \sin(0) \right)$$= \left( \frac{3\pi}{2} - \frac{1}{2}(-1) - (-1) \right) - \left( 0 - \frac{1}{2}(1) - 0 \right)$$= \left( \frac{3\pi}{2} + \frac{1}{2} + 1 \right) - \left( -\frac{1}{2} \right)$$= \frac{3\pi}{2} + \frac{3}{2} + \frac{1}{2} = \frac{3\pi}{2} + 2$.

The area enclosed by the curves $y = \cos x$,$y = 1 + \sin 2x$ and $x = \frac{3\pi}{2}$ in the first and fourth quadrants (as shown in the figure) is:

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