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Chapter 4: Problem 86

Find the area of the region. Use a graphing utility to verify your result. $$ y=\sin x+\cos 2 x $$

Short Answer

Expert verified
The area of the region defined by the function \(y = \sin x + \cos2x\) over the interval [0, \(2\pi\)] is 0.

Step by step solution

01

Identifying the Interval

Looking at the function, it should be noticed that it is periodic, hence the area of one period is enough. Choosing the interval [0, \(2\pi\)] is a good option as it consider one period for both sine and cos function.
02

Applying the Integral

Now, the integral can be applied to the function over the interval [0, \(2\pi\)]. Using the formulas \(\int\sin x dx = -\cos x + C\) and \(\int\cos x dx = \sin x + C\), the area A is expressed as \(A = \int_{0}^{2\pi} (\sin x + \cos2x) dx\). The integral can be solved by splitting it into two parts, resulting in \(A = \int_{0}^{2\pi} \sin x dx + \int_{0}^{2\pi} \cos2x dx\). Then, applying the formulas gives \(-[\cos(2\pi) - \cos(0)] + [0.5*\sin(2*2\pi) - 0.5*\sin(2*0)]\).
03

Evaluating the Integral

Evaluating the resulting expression gives the area of the region. Evaluating \(-[\cos(2\pi) - \cos(0)] + [0.5*\sin(2*2\pi) - 0.5*\sin(2*0)]\), it's concluded that it equals 0.
04

Verifying the Result Using a Graphing Utility

Using a graphing utility, plot the function \(y = \sin x + \cos2x\) over the interval [0, \(2\pi\)]. The area under the curve within this interval should match the calculated result.

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