Definition of the Definite Integral — Question 3

Question 3 - Solution

Divide the interval $[0,4]$ into $n$ equal subintervals. Each subinterval has width $$\Delta x=\frac{4}{n}.$$

Using left endpoints, the $i$th sample point is $$x_{i-1}=(i-1)\Delta x=\frac{4(i-1)}{n}.$$

Evaluate the function: $$f(x_{i-1})=\sqrt{\frac{4(i-1)}{n}}=2\sqrt{\frac{i-1}{n}}.$$

The Riemann sum is $$\sum_{i=1}^{n} f(x_{i-1})\,\Delta x = \sum_{i=1}^{n} 2\sqrt{\frac{i-1}{n}}\cdot\frac{4}{n} = \frac{8}{n^{3/2}}\sum_{i=1}^{n}\sqrt{i-1}.$$

Thus, $$\int_{0}^{4}\sqrt{x}\,dx = \lim_{n\to\infty} \frac{8}{n^{3/2}} \sum_{i=1}^{n}\sqrt{i-1}.$$

As $n\to\infty$, this Riemann sum converges to the definite integral: $$\int_{0}^{4}\sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_{0}^{4} =\frac{16}{3}.$$

$$\boxed{\frac{16}{3}}$$