SOLUTION 12: Compute the area of the region enclosed by the graphs of the equations $y = \sin \sqrt{x}$ and $y=0$ on the interval $[0, \pi^2]$. Now see the given graph of the enclosed region.

Using vertical cross-sections to describe this region, we get that $$0 \le x \le \pi^2 \ \ and \ \ 0 \le y \le \sin \sqrt{x} \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{\pi^2 } (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{0}^{\pi^2} \sin \sqrt{x} \ dx }$$ Use a power" u-substitution. Let $u^2=x$ (or $u= \sqrt{x}$) so that $2u \ du = dx$. Then $$\displaystyle { \int \sin \sqrt{x} \ dx } = { 2 \int u \cdot \sin u \ du }$$ (Now use integration by parts. Let $w=u$ and $dv = \sin u \cdot du$ so that $dw = du$ and $v= -\cos u$.) $$= 2 \Big( -u \cos u - \displaystyle { \int - \cos u \ du } \Big)$$ $$= -2 u \cos u + 2 \displaystyle { \int \cos u \ du }$$ $$= -2 u \cos u + 2 \sin u + C$$ $$= -2 \sqrt{x} \cos \sqrt{x} + 2 \sin \sqrt{x} + C$$ Thus, continuing with the definite integral, we get $$\displaystyle { \int_{0}^{\pi^2} \sin \sqrt{x} \ dx } = (-2 \sqrt{x} \cos \sqrt{x} + 2 \sin \sqrt{x}) \Big\vert_{0}^{\pi^2}$$ $$= (-2 \sqrt{{\pi}^2} \cos \sqrt{{\pi}^2} + 2 \sin \sqrt{{\pi}^2} ) - (-2 \sqrt{0} \cos \sqrt{0} + 2 \sin \sqrt{0} )$$ $$= (-2 \pi \cos \pi + 2 \sin \pi ) - (-2 \sqrt{0} \cos 0 + 2 \sin 0 )$$ $$= (-2 \pi (-1) + 2(0)) - (-2(0)(1)+2(0))$$ $$= 2 \pi$$