Definition of the Definite Integral — Question 7

Question 7

Let $$f(x)=3-2x$$ on the interval $[0,1]$.

Using the definition of the definite integral, express $$\int_{0}^{1}(3-2x)\,dx$$ as the limit of a Riemann sum using left endpoints, and evaluate the limit.

Question 7 - Solution

Partition the interval $[0,1]$ into $n$ equal subintervals. Each subinterval has width $$\Delta x=\frac{1}{n}.$$

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

Evaluate the function at each sample point: $$f(x_{i-1})=3-2\left(\frac{i-1}{n}\right).$$

The Riemann sum is $$\sum_{i=1}^{n} f(x_{i-1})\,\Delta x = \frac{1}{n} \sum_{i=1}^{n} \left(3-\frac{2(i-1)}{n}\right).$$

Separate the sum: $$\frac{1}{n} \left( \sum_{i=1}^{n}3 -\frac{2}{n}\sum_{i=1}^{n}(i-1) \right).$$

Using $$\sum_{i=1}^{n}1=n, \qquad \sum_{i=1}^{n}(i-1)=\frac{n(n-1)}{2},$$ we obtain $$\frac{1}{n} \left( 3n-\frac{2}{n}\cdot\frac{n(n-1)}{2} \right) = 3-(n-1)\frac{1}{n}.$$

Thus, $$\int_{0}^{1}(3-2x)\,dx = \lim_{n\to\infty}\left(3-\frac{n-1}{n}\right)=2.$$

$$\boxed{2}$$