Chapter 6.3, Problem 27E

To determine

ToEvaluate:The integral ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}$ using the antiderivative by interpreting the integrand as the rate of change of a quantity.

Answer to Problem 27E

  ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}=\frac{\pi }{2}$

Explanation of Solution

Given:

The integral ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}$

Concept Used:

To evaluate the definite integral of $f$ from $a$ to any number $x$ . Simply compute $F\left(x\right)-F\left(a\right)$ , where $F$ is any antiderivative of $f$ .

That is

  ${\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}=F\left(x\right)-F\left(a\right)$

Calculation:

Given the integral ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}$

Since $\frac{1}{1+x^2}$ is the rate of change of the quantity $F\left(x\right)={\tan }^{-1}\left(x\right)$

That is

  $F\text{'}\left(x\right)=\frac{1}{1+x^2}$

Evaluating theintegral ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}$ , using the formula, ${\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}=F\left(x\right)-F\left(a\right)$

  $\begin{array}{c}{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(-1\right)\\ =\frac{\pi }{4}-\left(-\frac{\pi }{4}\right)\\ =\frac{\pi }{4}+\frac{\pi }{4}\\ =\frac{2\pi }{4}\end{array}$

  $=\frac{\pi }{2}$

Therefore, ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}=\frac{\pi }{2}$

Conclusion:

  ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2}dx}=\frac{\pi }{2}$