Question

Show that if \(f(x)\) has period \(p,\) the average value of \(f\) is the same over any interval of length \(p .\) Hint: Write \(\int_{a}^{a+p} f(x) d x\) as the sum of two integrals \((a \text { to } p, \text { and } p\) to \(a+p)\) and make the change of variable \(x=t+p\) in the second integral.

Short answer

The average value of \(f(x)\) over any interval of length \(p\) is the same due to the periodicity of \(f(x)\).

Step-by-step solution

Understand the Problem

Verify that for a function with period \(p\), the average value of the function remains the same over any interval of length \(p\).

Define Average Value

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by: \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]

Given Interval

Given an interval from \(a\) to \(a + p\), the length of the interval is \(p\). So, we aim to show: \[ \frac{1}{p} \int_{a}^{a+p} f(x) \, dx \]

Split the Integral

Split the integral \(\int_{a}^{a+p} f(x) \, dx\) into two parts: \[ \int_{a}^{p} f(x) \, dx + \int_{p}^{a+p} f(x) \, dx \]

Change of Variable

In the second integral \(\int_{p}^{a+p} f(x) \, dx\), perform the change of variable \(x = t + p\). Letting \(dx = dt\), the integral becomes: \[ \int_{p}^{a+p} f(x) \, dx = \int_{0}^{a} f(t + p) \, dt \]

Use Periodicity

Since \(f(x)\) is periodic with period \(p\), we have \(f(t + p) = f(t)\). Substituting this into the integral: \[ \int_{0}^{a} f(t + p) \, dt = \int_{0}^{a} f(t) \, dt \]

Combine Integrals

Now, the full integral from \(a\) to \(a + p\) becomes: \[ \int_{a}^{a+p} f(x) \, dx = \int_{a}^{p} f(x) \, dx + \int_{0}^{a} f(t) \, dt \]

Evaluate Average

Since the intervals cover a period length, we can pair up parts to get the full average over any \(p\)-length interval: \[ \frac{1}{p} \int_{a}^{a+p} f(x) \, dx = \frac{1}{p} \int_{0}^{p} f(x) \, dx \]

Key concepts

Average Value

The average value of a function is a critical concept in integral calculus. It essentially gives us a single value that represents the average output of the function over a specific interval.
For a continuous function, the average value over an interval \([a, b]\) can be calculated using:

• \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]

This formula comes from summing up all the values of the function within the interval and then dividing by the length of the interval.

In our problem, we want to show that the average value of a periodic function stays the same over any interval of the same length. This is crucial for understanding how periodic functions behave across repeated cycles.

Integral Calculus

Integral calculus is the branch of calculus that deals with integrals and their properties.
In this problem, we use integrals to find areas under curves and to compute the average value of a function.

The integral of a function over an interval \([a, b]\) gives us the accumulated area under the curve of that function between those two points:

• \[ \int_{a}^{b} f(x) \, dx \]

Integrals are particularly helpful when dealing with functions that are not constant, as they allow us to sum up an infinite number of infinitesimally small contributions to get a total value.
This property is used in our exercise to split and re-combine integrals, showing the uniformity of the average value across intervals of the same length.

Periodicity

Periodicity refers to the property of a function repeating its values at regular intervals, known as periods. For a function \ f(x) \, if

• \[ f(x + p) = f(x) \] for all x,

then \ p \ is called the period of the function.

In the given problem, periodicity is a key aspect because it allows us to use the function’s behavior over one interval to infer its behavior over any other interval of the same length.
This is why the substitution and transformation of variables work effectively.
Periodic functions are seen frequently in real-world scenarios, such as seasonal temperatures, sound waves, and electromagnetic waves.

Change of Variables

The technique of changing variables is an important method in solving integrals, simplifying problems, and transforming intervals.
It involves substituting a new variable in place of the existing one to make the calculation or algebra simpler.

For example, consider the substitution

• \[ x = t + p \]

When changing variables, it's also necessary to adjust the differential element accordingly. In our example,

• \[ dx = dt \]

This method is particularly useful when dealing with periodic functions, as it helps to re-establish the starting interval to match periods accurately.
By applying this technique, we transform the function within the integral to leverage periodic behavior effectively.