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Find the area bounded by the region $y=x \sin(x)$, and $y=x$, for $0\le x\le \frac{\pi}{2}$.

My attempt

Area $=\int_\limits{0}^{\frac{\pi}{2}}(x-x\sin(x))dx$

After integrating I got:

$$[\frac{x^2}{2}+x\cos(x)-\sin(x)]_0^\frac{\pi}{2}$$

Is my answer right?

Which leads me to get approximately .2337 units squared.

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  • $\begingroup$ Why do you think this is wrong? $\endgroup$ – Peter Foreman Apr 22 at 13:06
  • $\begingroup$ @PeterForeman Because I'm new at doing this. $\endgroup$ – EnlightenedFunky Apr 22 at 13:07
  • $\begingroup$ You don't need to say "units squared". Otherwise it's right $\endgroup$ – man and laptop Apr 22 at 13:08
  • $\begingroup$ That’s correct, though I might show the exact answer $\pi^2/8 - 1$ before the approximation $\endgroup$ – MPW Apr 22 at 13:20
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Your answer is correct to four decimal places (see e.g. WolframAlpha for confirmation), just make sure you can also get the correct exact answer (in terms of $\pi$).

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  • $\begingroup$ Is there a way to do that for volume? $\endgroup$ – EnlightenedFunky Apr 22 at 13:10
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    $\begingroup$ Do you mean find the volume between two 3-dimensional curves? By the way, you can also rotate a 2-dimensional area around a line, or you can make squares/semicircles/triangles which have the 'height' of the curve as their base. $\endgroup$ – Toby Mak Apr 22 at 13:13
  • $\begingroup$ @TobyMak No on WolframAlpha to do volume $\endgroup$ – EnlightenedFunky Apr 22 at 13:17
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    $\begingroup$ Sorry, I still don't get what you mean. Can you explain further? $\endgroup$ – Toby Mak Apr 22 at 13:19
  • $\begingroup$ @TobyMak wolframalpha.com/input/?i=volume+of+a+x%5E2+from+0+%3C+x+%3C+2 to do this I found it thanks by the way. $\endgroup$ – EnlightenedFunky Apr 22 at 13:20

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