Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 329: 68

$\int_{0}^1 \sqrt{x}dx=\frac{2}{3}$

$$\begin{align*} \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n}\sqrt{\frac{i}{n}}&=\lim_{n \to \infty} \sum_{i=1}^{n}\sqrt{0+\frac{i}{n}}\frac{1-0}{n}\\ &=\lim_{n \to \infty} \sum_{i=1}^{n}\sqrt{x_i}\frac{1-0}{n}\\ &=\int_{0}^1 \sqrt{x}dx\\ &=\left[\frac{2}{3}\sqrt{x^3}\right]_0^1\\ &=\frac{2}{3} \end{align*}$$