Determine the coordinates of the centroid $(\bar{x},\bar{y})$ of the region bounded by $y=\ln x$, the $x$‑axis, $x=1$ and $x=e^2$. Give your answer in exact form.

Step-by-step solution

1. The area of the region is $A={\int }_1^{e^2}\ln xdx=[x\ln x-x{]}_1^{e^2}=e^2+1$.
2. For $\bar{x}$, compute ${\int }_1^{e^2}x\ln xdx$ using integration by parts: $\int x\ln xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}$.
3. Evaluating gives $\left(\frac{e^4}{2}\cdot 2-\frac{e^4}{4}\right)-\left(\frac{1}{2}\cdot 0-\frac{1}{4}\right)=\frac{3}{4}{e}^4+\frac{1}{4}$.
4. Hence $\bar{x}=\frac{1}{A}\left(\frac{3}{4}{e}^4+\frac{1}{4}\right)=\frac{\frac{3}{4}{e}^4+\frac{1}{4}}{e^2+1}$.
5. For $\bar{y}$, use $\bar{y}=\frac{1}{2A}{\int }_1^{e^2}(\ln x{)}^{2}dx$.
6. Integral: $\int (\ln x{)}^{2}dx=x[(\ln x{)}^2-2\ln x+2]$.
7. Evaluating from $1$ to $e^2$ gives $2e^2-2=2(e^2-1)$.
8. Thus $\bar{y}=\frac{1}{2A}\cdot 2(e^2-1)=\frac{e^2-1}{e^2+1}$.
9. The centroid is $(\frac{\frac{3}{4}{e}^4+\frac{1}{4}}{e^2+1},\frac{e^2-1}{e^2+1})$.

Concept tested: Centroid of a plane region

Published 18 September 2026 · How this solution was prepared and checked