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Chapter 5: Problem 5

$$\text { Use graphs to evaluate } \int_{0}^{2 \pi} \sin x d x \text { and } \int_{0}^{2 \pi} \cos x d x$$

Short Answer

Expert verified
Answer: The definite integrals \(\int_{0}^{2 \pi} \sin x dx = 0\) and \(\int_{0}^{2 \pi} \cos x dx = 0\).

Step by step solution

01

Sketch the graph of the sine function

On the interval \([0, 2\pi]\), plot the sine function. Observe the graph goes from \(0\) to \(1\), back to \(0\), then down to \(-1\), and back up to \(0\) again. The graph of the sine function is symmetric with respect to the vertical line \(x = \pi\).
02

Identify and split the area under the curve

Note that on the interval \([0, \pi]\), the sine function is positive, while it's negative on the interval \((\pi, 2\pi]\). Therefore, the integral \(\int_{0}^{2 \pi} \sin x dx\) equals the sum of areas under the curve in the interval \([0, \pi]\) minus the area under the curve in the interval \([\pi, 2 \pi]\). However, due to symmetry, these two areas will cancel each other out, and the integral will equal zero.
03

Conclusion

Therefore, \(\int_{0}^{2 \pi} \sin x dx = 0\) #Part 2: Evaluate \(\int_{0}^{2 \pi} \cos x dx\) # Now we will proceed with steps to evaluate the definite integral of the cosine function between \(0\) and \(2\pi\).
04

Sketch the graph of the cosine function

On the interval \([0, 2\pi]\), plot the cosine function. Observe the graph goes from \(1\) to \(0\), down to \(-1\), back to \(0\), and then back up to \(1\). The graph of the cosine function is also symmetric with respect to the vertical line \(x = \pi\).
05

Identify and split the area under the curve

Note that on the interval \([0, \pi]\), the cosine function is positive in the interval \((0, \frac{\pi}{2}]\) and negative in the interval \([\frac{\pi}{2}, \pi]\), with the integral representing the sum of the areas under the curve for those two intervals. Similarly, in the interval \([\pi, 2 \pi]\), the cosine function is positive for the interval \([\frac{3\pi}{2}, 2\pi]\), and negative for interval \([\pi, \frac{3\pi}{2}]\). Thus, the integral equals the difference of the areas under the curve in each interval, but due to symmetry, these two areas will also cancel each other out.
06

Conclusion

Therefore, \(\int_{0}^{2 \pi} \cos x dx = 0\) In conclusion, both integrals equal zero due to the symmetry of the sine and cosine functions.

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