Area — Question 3

Question 3 - Solution

First determine which curve lies above the other on $[0,1]$. For $0<x<1$, $$\sqrt{x} \ge x^2.$$

Thus, the enclosed area is $$A=\int_{0}^{1}\bigl(\sqrt{x}-x^2\bigr)\,dx.$$

Evaluate each term: $$\int_{0}^{1}\sqrt{x}\,dx =\left[\frac{2}{3}x^{3/2}\right]_0^1 =\frac{2}{3},$$ $$\int_{0}^{1}x^2\,dx =\left[\frac{x^3}{3}\right]_0^1 =\frac{1}{3}.$$

Subtract: $$A=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}.$$

$$\boxed{\frac{1}{3}}$$