Definition of the Definite Integral — Question 8

Question 8 - Solution

Partition the interval $[1,e]$ into $n$ equal subintervals. The width of each subinterval is $$\Delta x=\frac{e-1}{n}.$$

Using right endpoints, the $i$th sample point is $$x_i=1+i\Delta x=1+i\frac{e-1}{n}.$$

Evaluate the function at each sample point: $$f(x_i)=\frac{1}{1+i\frac{e-1}{n}}.$$

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

Thus, $$\int_{1}^{e}\frac{1}{x}\,dx = \lim_{n\to\infty} \frac{e-1}{n} \sum_{i=1}^{n} \frac{1}{1+i\frac{e-1}{n}}.$$

Evaluating the definite integral, $$\int_{1}^{e}\frac{1}{x}\,dx = \bigl[\ln x\bigr]_{1}^{e} = 1.$$

$$\boxed{1}$$