Find the area enclosed between the curves $y=x^3$ and $y=3x^2$ between their intersection points.

Step-by-step solution

1. Find intersection points by solving $x^3=3x^2$: $x^2(x-3)=0$ gives $x=0$ or $x=3$.
2. Determine which curve lies above the other between $0$ and $3$ by evaluating at $x=1$: $3(1{)}^2=3$ and $1^3=1$, so $3x^2$ is above $x^3$.
3. Set up the area integral: $A={\int }_0^3(3x^2-x^3),dx$.
4. Integrate: $\int 3x^2,dx=x^3$ and $\int x^3,dx=\frac{1}{4}{x}^4$.
5. Compute $A=[x^3-\frac{1}{4}{x}^4{]}_0^3=27-\frac{1}{4}cdot81=27-20.25=6.75$.
6. Express $6.75$ as a fraction: $6.75=\frac{27}{4}$.
7. Thus the area of the region is $\frac{27}{4}$ square units.

Concept tested: Area between curves

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked