Definition of the Definite Integral — Question 4

Question 4

Let $$f(x)=\frac{1}{1+x^2}$$ on the interval $[-1,1]$.

Using the definition of the definite integral, express $$\int_{-1}^{1}\frac{1}{1+x^2}\,dx$$ as the limit of a Riemann sum using right endpoints, and evaluate the limit.

Question 4 - Solution

Divide the interval $[-1,1]$ into $n$ equal subintervals. The width of each subinterval is $$\Delta x=\frac{1-(-1)}{n}=\frac{2}{n}.$$

Using right endpoints, the $i$th sample point is $$x_i=-1+i\Delta x=-1+\frac{2i}{n}.$$

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

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

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

Evaluating the definite integral, $$\int_{-1}^{1}\frac{1}{1+x^2}\,dx = \bigl[\arctan x\bigr]_{-1}^{1} = \frac{\pi}{4}-\left(-\frac{\pi}{4}\right) = \frac{\pi}{2}.$$

$$\boxed{\frac{\pi}{2}}$$