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

Step-by-step solution

1. The area of the region is $A={\int }_1^{e^2}\ln xdx$. Using integration by parts, $\int \ln xdx=x\ln x-x$, so $A=[x\ln x-x{]}_1^{e^2}=e^2\cdot 2-e^2-(0-1)=e^2+1$.
2. The $x$‑coordinate of the centroid is $\bar{x}=\frac{1}{A}{\int }_1^{e^2}x\ln xdx$. Again by parts, $\int x\ln xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}$. Evaluating gives $\left(\frac{e^4}{2}\cdot 2-\frac{e^4}{4}\right)-\left(\frac{1}{2}\cdot 0-\frac{1}{4}\right)=\frac{3e^4+1}{4}$. Hence $\bar{x}=\frac{\frac{3e^4+1}{4}}{e^2+1}=\frac{3e^4+1}{4(e^2+1)}$.
3. The $y$‑coordinate is $\bar{y}=\frac{1}{2A}{\int }_1^{e^2}(\ln x{)}^{2}dx$. Using the formula $\int (\ln x{)}^{2}dx=x(\ln x{)}^2-2x\ln x+2x$, we get the definite integral $[x(\ln x{)}^2-2x\ln x+2x{]}_1^{e^2}=2e^2-2=2(e^2-1)$. Therefore $\bar{y}=\frac{1}{2(e^2+1)}\cdot 2(e^2-1)=\frac{e^2-1}{e^2+1}$.
4. Thus the centroid is $\left(\frac{3e^4+1}{4(e^2+1)},\frac{e^2-1}{e^2+1}\right)$.

Concept tested: Centroid of a plane region

Published 11 September 2026 · How this solution was prepared and checked