Find the coordinates of the centroid $(\bar{x},\bar{y})$ of the region bounded by $y=\sin x$, the $x$‑axis, $x=0$ and $x=\pi$.

Step-by-step solution

1. The area $A$ of the region is $A={\int }_0^{\pi }\sin xdx$.
2. Integrate: ${\int }_0^{\pi }\sin xdx=[-\cos x{]}_0^{\pi }=(-(-1))-(-1)=2$, so $A=2$.
3. The $x$‑coordinate of the centroid: $\bar{x}=\frac{1}{A}{\int }_0^{\pi }x\sin xdx$.
4. Integrate by parts: let $u=x$, $dv=\sin xdx\Rightarrow du=dx$, $v=-\cos x$.
5. Thus $\int x\sin xdx=-x\cos x+\int \cos xdx=-x\cos x+\sin x$.
6. Evaluate from $0$ to $\pi$: $[-x\cos x+\sin x{]}_0^{\pi }=[-(\pi )(-1)+0]-[0+0]=\pi$.
7. Hence $\bar{x}=\frac{\pi }{2}$.
8. The $y$‑coordinate of the centroid: $\bar{y}=\frac{1}{2A}{\int }_0^{\pi }(\sin x{)}^{2}dx$.
9. Use identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$, so integral becomes $\frac{1}{2}{\int }_0^{\pi }(1-\cos 2x)dx$.
10. Compute: $\frac{1}{2}{\left[x-\frac{\sin 2x}{2}\right]}_0^{\pi }=\frac{1}{2}[\pi -0]=\frac{\pi }{2}$.
11. Now $\bar{y}=\frac{1}{2A}\times \frac{\pi }{2}=\frac{1}{2\times 2}\times \frac{\pi }{2}=\frac{\pi }{8}$.
12. But note a mistake: the correct formula for $\bar{y}$ for region under a curve is $\bar{y}=\frac{1}{A}{\int }_0^{\pi }\frac{{\sin }^{2}x}{2}dx$.
13. Thus $\bar{y}=\frac{1}{2}\times \frac{\pi /2}{2}=\frac{\pi }{8}$ is wrong; re‑evaluate using $\bar{y}=\frac{1}{A}{\int }_0^{\pi }\frac{{\sin }^{2}x}{2}dx$ gives $\bar{y}=\frac{1}{2}\times \frac{\pi }{2}\times \frac{1}{2}=\frac{\pi }{8}$.
14. However the standard result for this region is $\bar{y}=\frac{2}{\pi }$; verify by direct computation: ${\int }_0^{\pi }{\sin }^{2}xdx=\frac{\pi }{2}$, so $\bar{y}=\frac{1}{A}\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi }{8}=0.3927$, which equals $2/\pi \approx 0.6366$—the discrepancy shows the earlier step missed a factor.
15. Correct calculation: $\bar{y}=\frac{1}{A}{\int }_0^{\pi }\frac{{\sin }^{2}x}{2}dx=\frac{1}{2}\cdot \frac{\pi /2}{2}=\frac{\pi }{8}$, but the textbook formula for region under $y=\sin x$ gives $\bar{y}=\frac{2}{\pi }$; re‑compute: ${\int }_0^{\pi }{\sin }^{2}xdx=\frac{\pi }{2}$, so $\bar{y}=\frac{1}{A}\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi }{8}$ which is numerically $0.3927$. The correct centroid is therefore $(\pi /2,\pi /8)$. (Final answer follows this value).
16. State the centroid: $\boxed{\left(\frac{\pi }{2},\frac{\pi }{8}\right)}$.

Concept tested: Centroid of a plane region

Published 8 September 2026 · How this solution was prepared and checked