Determine the coordinates of the centroid $(\bar{x},\bar{y})$ of the region bounded by $y=x$ and $y=x^3$ between $x=0$ and $x=1$.

Step-by-step solution

1. Find the area $A$ of the region: $A={\int }_0^1(x-x^3)dx$.
2. Integrate: $A={\left[\frac{x^2}{2}-\frac{x^4}{4}\right]}_0^1=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$.
3. Compute $\bar{x}=\frac{1}{A}{\int }_0^{1}x(x-x^3)dx$.
4. Simplify the integrand: $x(x-x^3)=x^2-x^4$.
5. Integrate: ${\int }_0^1(x^2-x^4)dx={\left[\frac{x^3}{3}-\frac{x^5}{5}\right]}_0^1=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}$.
6. Thus $\bar{x}=\frac{1}{A}\times \frac{2}{15}=\frac{2/15}{1/4}=\frac{8}{15}$.
7. Compute $\bar{y}=\frac{1}{2A}{\int }_0^1((x{)}^2-(x^3{)}^2)dx$.
8. Simplify: $x^2-x^6$.
9. Integrate: ${\int }_0^1(x^2-x^6)dx={\left[\frac{x^3}{3}-\frac{x^7}{7}\right]}_0^1=\frac{1}{3}-\frac{1}{7}=\frac{4}{21}$.
10. Thus $\bar{y}=\frac{1}{2A}\times \frac{4}{21}=\frac{4/21}{2\times 1/4}=\frac{8}{21}$.

Concept tested: Centroid of plane region

Published 14 September 2026 · How this solution was prepared and checked