Chapter 6.3, Problem 19E

To determine

ToEvaluate:The integral ${\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\sin \left(x\right)dx}$ using the antiderivative by interpreting the integrand as the rate of change of a quantity.

Answer to Problem 19E

  ${\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\sin \left(x\right)dx}=-2$

Explanation of Solution

Given:

The integral ${\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\sin \left(x\right)dx}$

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 ${\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\sin \left(x\right)dx}$

Since $\sin \left(x\right)$ is the rate of change of the quantity $F\left(x\right)=-\cos \left(x\right)$

That is

  $\begin{array}{c}F\text{'}\left(x\right)=-\left(-\sin \left(x\right)\right)\\ =\sin \left(x\right)\end{array}$

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

  $\begin{array}{c}{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\sin \left(x\right)dx}=\left(-\cos \left(2\pi \right)\right)-\left(-\cos \left(\pi \right)\right)\\ =-1+\cos \left(\pi \right)\text{ Since, }\cos \left(\pi \right)=1\\ =-1-1\text{ Since, }\cos \left(\pi \right)=-1\\ =-2\end{array}$

Therefore, ${\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\sin \left(x\right)dx}=-2$

Conclusion:

  ${\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\sin \left(x\right)dx}=-2$