Area Solution 12

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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.

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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

$\displaystyle { \int_{0}^{\pi^2} \sin \sqrt{x} \ dx } = (-2 \sqrt{x} \cos \sqrt{x} + 2 \sin \sqrt{x}) \Big\vert_{0}^{\pi^2}$

$= (-2 \sqrt{{\pi}^2} \cos \sqrt{{\pi}^2} + 2 \sin \sqrt{{\pi}^2} ) - (-2 \sqrt{0} \cos \sqrt{0} + 2 \sin \sqrt{0} )$

$= (-2 \pi \cos \pi + 2 \sin \pi ) - (-2 \sqrt{0} \cos 0 + 2 \sin 0 )$

$= (-2 \pi (-1) + 2(0)) - (-2(0)(1)+2(0))$

$= 2 \pi$

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