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Chapter 4: Problem 85

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

Short Answer

Expert verified
The area under the curve \(y = 2\sin(x) + \sin(2x)\) from \(0\) to \(2\pi\) is equal to \(0\).

Step by step solution

01

Identify the Integral Boundaries

The integral boundaries are determined by the period of the function, which is here \(2\pi\). So the boundaries will range from 0 to \(2\pi\).
02

Set Up the Integral

The area under the curve is found by integrating: \(\int_a^b f(x) \,dx\). Set up the integral for given function \(y = 2\sin(x) + \sin(2x)\) from 0 to \(2\pi\). So, the equation becomes: \(\int_0^{2\pi} [2\sin(x) + \sin(2x)] \,dx\).
03

Evaluate the Integral

Evaluate the integral. This involves integrating each of the parts separately. The antiderivative of \(2\sin(x)\) is \(-2\cos(x)\) and the antiderivative of \(\sin(2x)\) is \(-\frac{1}{2}\cos(2x)\). After evaluating these at \(2\pi\) and subtracting the value at \(0\), get the final result.

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Most popular questions from this chapter

In Exercises \(69-74\), find the indefinite integral using the formulas of Theorem 4.24 \(\int \frac{1}{\sqrt{1+e^{2 x}}} d x\)

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