Find the region bounded by $y=x \sin x$, and $y=x$

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:

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

Which leads me to get approximately .2337 units squared.

• Why do you think this is wrong? – Peter Foreman Apr 22 at 13:06
• @PeterForeman Because I'm new at doing this. – EnlightenedFunky Apr 22 at 13:07
• You don't need to say "units squared". Otherwise it's right – man and laptop Apr 22 at 13:08
• That’s correct, though I might show the exact answer $\pi^2/8 - 1$ before the approximation – MPW Apr 22 at 13:20

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 $$\pi$$).