The area between the curve y = sin^2 x,the x- | vedclass

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(B) The required area $A$ is given by the integral of the function $y = \sin^2 x$ from $x = 0$ to $x = \frac{\pi}{2}$.$A = \int_0^{\pi/2} \sin^2 x \,

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$\frac{\pi}{4}$

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(B) The required area $A$ is given by the integral of the function $y = \sin^2 x$ from $x = 0$ to $x = \frac{\pi}{2}$.$A = \int_0^{\pi/2} \sin^2 x \, dx$Using the trigonometric identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we get:$A = \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx$$A = \frac{1}{2} \int_0^{\pi/2} (1 - \cos 2x) \, dx$$A = \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_0^{\pi/2}$$A = \frac{1}{2} \left( (\frac{\pi}{2} - \frac{\sin \pi}{2}) - (0 - \frac{\sin 0}{2}) \right)$$A = \frac{1}{2} (\frac{\pi}{2} - 0 - 0 + 0) = \frac{\pi}{4}$Thus,the area is $\frac{\pi}{4}$ square units.

The area between the curve $y = \sin^2 x$,the $x$-axis,and the ordinates $x = 0$ and $x = \frac{\pi}{2}$ is:

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