Question

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

Short answer

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

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.

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 \(-[\cos(2\pi) - \cos(0)] + [0.5*\sin(2*2\pi) - 0.5*\sin(2*0)]\).

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.

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.