Definition of the Definite Integral — Question 6

Question 6 - Solution

Divide the interval $[-2,2]$ into $n$ equal subintervals. Each subinterval has width $$\Delta x=\frac{2-(-2)}{n}=\frac{4}{n}.$$

Using right endpoints, the $i$th sample point is $$x_i=-2+i\Delta x=-2+\frac{4i}{n}.$$

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

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

Thus, $$\int_{-2}^{2}|x|\,dx = \lim_{n\to\infty} \frac{4}{n} \sum_{i=1}^{n} \left|-2+\frac{4i}{n}\right|.$$

To evaluate the integral, use symmetry: $$\int_{-2}^{2}|x|\,dx = 2\int_{0}^{2}x\,dx.$$

Compute: $$2\left[\frac{x^2}{2}\right]_{0}^{2}=4.$$

$$\boxed{4}$$