Definition of the Definite Integral — Question 9

Question 9

Let $$f(x)= \begin{cases} 2x, & 0\le x\le 1,\\[4pt] 2, & 1<x\le 2. \end{cases}$$

Using the definition of the definite integral, express $$\int_{0}^{2} f(x)\,dx$$ as the limit of a Riemann sum, and evaluate the limit.

Question 9 - Solution

Partition 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 $$x_i=\frac{2i}{n}.$$

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

Split the sum according to where the function changes behavior. The condition $x_i\le 1$ corresponds to $$\frac{2i}{n}\le 1 \quad\Rightarrow\quad i\le \frac{n}{2}.$$

Thus, $$\sum_{i=1}^{n} f\!\left(\frac{2i}{n}\right) = \sum_{i=1}^{n/2} 2\left(\frac{2i}{n}\right) + \sum_{i=n/2+1}^{n} 2.$$

Compute each sum: $$\sum_{i=1}^{n/2} \frac{4i}{n} = \frac{4}{n}\cdot\frac{(n/2)(n/2+1)}{2} = \frac{n}{2}+1,$$ $$\sum_{i=n/2+1}^{n} 2 = 2\cdot\frac{n}{2} = n.$$

Therefore, $$\sum_{i=1}^{n} f\!\left(\frac{2i}{n}\right) = \frac{n}{2}+1+n = \frac{3n}{2}+1.$$

Multiply by $\Delta x$: $$\frac{2}{n}\left(\frac{3n}{2}+1\right) = 3+\frac{2}{n}.$$

Taking the limit, $$\int_{0}^{2} f(x)\,dx = \lim_{n\to\infty}\left(3+\frac{2}{n}\right) = 3.$$

$$\boxed{3}$$