The area bounded by the curve y=sin^{2} x,the | vedclass

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

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$\frac{\pi}{4}$ sq. units

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(C) The required area $A$ is given by the definite integral of the function $y = \sin^{2} x$ from $x = 0$ to $x = \frac{\pi}{2}$.$A = \int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx$Using the trigonometric identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we get:$A = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx$$A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx$$A = \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$$A = \frac{1}{2} \left[ \left( \frac{\pi}{2} - \frac{\sin \pi}{2} \right) - (0 - 0) \right]$$A = \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{4}$ sq. units.

The area bounded by the curve $y=\sin^{2} x$,the $x$-axis,and the lines $x=0$ and $x=\frac{\pi}{2}$ is

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