Question

Find the average value of the function on the given interval. Use equation (4.8) if it applies. If an average value is zero, you may be able to decide this from a quick sketch which shows you that the areas above and below the \(x\) axis are the same. $$\cos x \text { on }(0,3 \pi)$$

Short answer

The average value of \f\text{cos}\f x\f on (0, 3π) is 0.

Step-by-step solution

Recall the Average Value Formula

The average value of a function \f(x)\f on an interval \f[a, b]\f is given by \[\frac{1}{b-a} \times \text{\textintegral\text{ }}_{a}^{b}f(x)\text{ }dx\].Here, the function is \f\text{cos}\f x\f and the interval is (0, 3π).

Set Up the Integral

Set up the integral for the average value:\[\frac{1}{3\text{π}-0} \times \text{\textintegral\text{ }}_{0}^{3\text{π}}\text{cos}\f(x)\text{ }dx\].

Compute the Integral

Solve the integral. The antiderivative of \f\text{cos}\f x\f is \f\text{sin}\f x\f, so:\[\text{\textintegral\text{ }}_{0}^{3\text{π}}\text{cos}\f(x)\text{ }dx = \text{sin}(x) \bigg|_{0}^{3\text{π}}= \text{sin}(3\text{π}) - \text{sin}(0)= 0 - 0= 0\].

Calculate the Average Value

Now plug the integral result back into the average value formula:\[\frac{1}{3\text{π}} \times 0 = 0\].

Key concepts

Definite Integral

A definite integral is a mathematical concept used to calculate the area under a curve over a specific interval. When dealing with the function \(f(x)\), the definite integral from \(a\) to \(b\) is written as \[ \text{\textintegral{}}_{a}^{b}f(x)dx \].
The limits \(a\) and \(b\) represent the interval's endpoints on the x-axis.
By providing these endpoints, we can calculate the cumulative value (or total area) between these points.
Integrals are used in various fields, such as physics, engineering, and economics, to measure quantities like area, volume, and growth.
It's important to understand how to compute an integral, which involves finding the antiderivative of the given function. In this exercise, the function is \(\text{cos} x\), and we need to integrate it from \(0\) to \(3\text{π}\).

Cosine Function

The cosine function, denoted as \(\text{cos} x\), is a fundamental trigonometric function that describes the relationship between the angle \(x\) and the adjacent side divided by the hypotenuse of a right-angled triangle.
It is a periodic function with a period of \(2\text{π}\), meaning it repeats its values every \(2\text{π}\) units.
Its range is between \(-1\) and \(1\).
When we integrate the cosine function, we're looking for the area under the cosine curve within a specified interval.
In our case, we integrate \( \text{cos} x \) from \(0\) to \(3\text{π}\). The antiderivative of \( \text{cos} x \) is \( \text{sin} x \), which we use to evaluate the integral.
This periodic behavior also means that areas above the x-axis and below it might cancel each other out over one period.

Average Value Formula

The average value of a continuous function \(f(x)\) over an interval \( [a, b]\) is determined using the average value formula: \[ \frac{1}{b-a} \times \text{\textintegral{}}_{a}^{b} f(x)dx \].
This formula calculates the mean (average) value of the function across that specific interval.
It combines both the concept of the definite integral and the length of the interval. In the given exercise, our task is to find the average value of \( \text{cos} x \) over the interval \([0, 3\text{π}]\).
Following this formula, we'll compute the integral \(\text{\textintegral{}}_{0}^{3\text{π}} \text{cos} x \) and then divide it by the length of the interval, \(3\text{π}\).
By solving the integral and applying the formula, we discover that the average value of \( \text{cos} x \) on \([0, 3\text{π}]\) is \(0\).
This result is often intuitive because the positive and negative areas under one complete period of the cosine function cancel each other out, leading to an average of zero.