Find the area of the region enclosed between the curves $y=2x$ and $y=x^2$ from $x=0$ to $x=2$.

Step-by-step solution

1. Identify the intersection points: solve $2x=x^2$ to get $x=0$ or $x=2$.
2. Since $2xge{x}^2$ on $[0,2]$, the top curve is $y=2x$ and the bottom is $y=x^2$.
3. Set up the integral for the area: $A={\int }_0^2(2x-x^2),dx$.
4. Integrate: $\int (2x-x^2),dx=x^2-\frac{x^3}{3}{|}_0^2$.
5. Evaluate: $A=(4-\frac{8}{3})-0=\frac{12}{3}-\frac{8}{3}=\frac{4}{3}$.
6. Realise the mistake: the computed value $4/3$ is the area under $2x-x^2$; but the integral of $2x-x^2$ actually gives $4/3$.
7. Re‑check: $x^2-\frac{x^3}{3}$ at $x=2$ gives $4-\frac{8}{3}=\frac{12-8}{3}=\frac{4}{3}$. So area is $\frac{4}{3}$.
8. But the area asked is $\frac{4}{3}$ units${}^2$.

Concept tested: Area between curves

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked