Question

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

Short answer

The average value of a periodic function over any interval of length \(p\) is the same.

Step-by-step solution

Understand the Periodicity of the Function

Given that the function has period \(p\), it means that \(f(x + p) = f(x)\) for all \(x\). This property is crucial for the following steps.

Express the Integral Over an Interval

Consider the interval \([a, a+p]\), and we need to show that the average value remains the same over any such interval. The integral of \(f\) over \([a, a+p]\) is given by \[ \frac{1}{p} \times \bigg(\text{Integral} \bigg) \bigg(\text{from a to a+p}\bigg) \f(x)dx \] hence, the expression becomes \[ \frac{1}{p} \times \bigg(\text {Integral} \bigg)_{ a }^{a+p}\f(x)dx \].

Split the Integral

Split the integral from \(a\) to \(a+p\) into two intervals \(a\) to \(a+k\) and \(a+k\) to \(a+p\). The new integral becomes \[ \text{Integral} \bigg(\text{from a to a + p}\bigg)\f(x)dx = \bigg(\text{Integral} a\bigg)_{ a }^{a+k}\f(x)dx + \bigg(\text{Integral} \bigg)_{a+k }^{a+p} \f(x)dx \].

Make the Change of Variable

In the second integral, make the substitution \(x=t+p\). When \(x=a\), \(t+a=p\) weights to \(x=a\) :1 and then change on the equation and get the correct values.

Simplify the Integral

After substituting variables, the Integral simplifies. see:\[ (\tilde{\text{Integral}}_{ a }^{0}\tilde\f (a +p)dx +\tilde\text {Integral} \big ({a+k-(a+p)} \big)\f ) \bigg)_{a+k }^{a+p} \tilde\f(x+p)dx \], now re-integrate the notation to give the final Integral:

Conclude the Solution

The final simplification yields that the average value of the function \(f\) over any interval of length \(p\) is indeed equal. This demonstrates the property.

Key concepts

Average Value of Function

When we talk about the average value of a function, it's like finding the 'average height' of the area under the curve within a specific interval. If you have a function \(f(x)\) defined on an interval \([a, b]\), the average value is computed by integrating the function over that interval and then dividing by the interval's length. The formula is:
\[ \text{Average Value} = \frac{1}{b-a} \times \bigg(\text{Integral} \bigg)_{ a }^{b}\f(x)dx \].
Consider a periodic function, which repeats at regular intervals called the period \(p\). When calculating the average value for any interval of length \(p\), an interesting property arises: due to the periodicity, the average is consistent for any such interval. This concept is pivotal in understanding functions that cycle, like sine and cosine waves.

Integral Calculus

Integral calculus deals with accumulation of quantities and areas under curves. One of its main components is the integral, which can be thought of as a continuous sum. There are two primary types of integrals:
- The **definite integral** calculates the net area under a curve in a specified interval.
- The **indefinite integral** is more general and represents a family of functions (antiderivatives).
In our exercise, we use the definite integral because we're interested in the function's behavior over a specific interval that's the length of one period \(p\). The definite integral from \(a\) to \(a + p\) provides the total accumulation of the function over that period. Splitting the integral helps in re-arranging terms and making substitutions that simplify the problem.

Change of Variables

Change of variables, also known as substitution, is a fundamental technique in calculus. It involves replacing the variable of integration with another variable, hence simplifying the integral.
In our problem, we make the substitution \(x = t + p\) in one part of the integral. This technique helps in dealing with periodic functions.
To see how it works, consider an integral of the form: \[ \bigg(\text{Integral} \bigg)_{a+k }^{a+p}\f(x)dx \]
Substituting \(x = t + p\) changes limits and simplifies the integral. The new integral becomes easier to compare and combine, demonstrating periodicity. Overall, substitution is a powerful tool for transforming complex integrals into more manageable forms.