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(D) The area $A$ of the region bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral $A = \int_{a}^{

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$5$

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(D) The area $A$ of the region bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral $A = \int_{a}^{b} |f(x)| \, dx$. Here,$f(x) = 5 \sin x$,$a = 0$,and $b = \frac{\pi}{2}$. Since $\sin x \geq 0$ for $0 \leq x \leq \frac{\pi}{2}$,we have: $A = \int_{0}^{\frac{\pi}{2}} 5 \sin x \, dx$ $A = 5 [-\cos x]_{0}^{\frac{\pi}{2}}$ $A = 5 [-\cos(\frac{\pi}{2}) - (-\cos(0))]$ $A = 5 [0 + 1]$ $A = 5 	imes 1 = 5$ sq. units. Thus,the correct option is $D$.

The area of the region bounded by the curve $y = 5 \sin x$,the $X$-axis,and the lines $x = 0$ and $x = \frac{\pi}{2}$ is . . . . . . sq. units.

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