Definition of the Definite Integral — Question 1

Question 1

Consider the function $$f(x)=2x+1$$ on the interval $[0,2]$.

Using the definition of the definite integral, express $$\int_{0}^{2} (2x+1)\,dx$$ as the limit of a Riemann sum with $n$ equal subintervals.

Question 1 - Solution

Divide the interval $[0,2]$ into $n$ equal subintervals. Each subinterval has width $$\Delta x=\frac{2}{n}.$$

Using right endpoints, the $i$th sample point is $$x_i=0+i\Delta x=\frac{2i}{n}.$$

Evaluate the function at $x_i$: $$f(x_i)=2\left(\frac{2i}{n}\right)+1=\frac{4i}{n}+1.$$

The Riemann sum is $$\sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n}\left(\frac{4i}{n}+1\right)\frac{2}{n}.$$

Simplify: $$\frac{2}{n}\sum_{i=1}^{n}\left(\frac{4i}{n}+1\right) = \frac{8}{n^2}\sum_{i=1}^{n} i+\frac{2}{n}\sum_{i=1}^{n}1.$$

Using $$\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n}1=n,$$ we obtain $$\frac{8}{n^2}\cdot\frac{n(n+1)}{2}+\frac{2}{n}\cdot n = 4\left(1+\frac{1}{n}\right)+2.$$

Thus, $$\int_{0}^{2}(2x+1)\,dx = \lim_{n\to\infty}\left(6+\frac{4}{n}\right)=6.$$

$$\boxed{6}$$