AVERAGE VALUE OF A FUNCTION

4. AVERAGE VALUE OF A FUNCTION

The concept of the average value of a function is often useful. You know how to find the average of a set of numbers: you add them and divide by the number of numbers.

Figure 4.1

348 Fourier Series and Transforms Chapter 7

This process suggests that we ought to get an approximation to the average value of a function f (x) on the interval (a, b) by averaging a number of values of f (x) (Figure 4.1):

(4.1) Average of f (x) on (a, b) is approximately equal to

f (x 1 ) + f (x 2 ) + · · · + f(x n ) . n

This should become a better approximation as n increases. Let the points x 1 ,x 2 ,···

be ∆x apart. Multiply the numerator and the denominator of the approximate average by ∆x. Then (4.1) becomes:

(4.2) Average of f (x) on (a, b) is approximately equal to

[f (x 1 ) + · · · + f(x n )]∆x . n ∆x

Now n ∆x = b −a, the length of the interval over which we are averaging, no matter what n and ∆x are. If we let n → ∞ and ∆x → 0, the numerator approaches

a f (x) dx, and we have

b f (x) dx (4.3)

Average of f (x) on (a, b) = a .

b−a

In applications, it may happen that the average value of a given function is zero. Example 1. The average of sin x over any number of periods is zero. The average value of

the velocity of a simple harmonic oscillator over any number of vibrations is zero. In such cases the average of the square of the function may be of interest.

Example 2. If the alternating electric current flowing through a wire is described by a sine function, the square root of the average of the sine squared is known as the root-mean-square or effective value of the current, and is what you would measure with an a-c ammeter. In the example of the simple harmonic oscillator, the average

kinetic energy (average of 1 2 mv 2 ) is 1 2 m times the average of v 2 .

Figure 4.2

Now you can, of course, find the average value of sin 2 x over a period (say −π to π) by evaluating the integral in (4.3). There is an easier way. Look at the graphs of cos 2 x and sin 2 x (Figure 4.2). You can probably convince yourself that the area

Section 4 Average Value of a Function 349

under them is the same for any quarter-period from 0 to π/2, π/2 to π, etc. (Also see Problems 2 and 13.) Then

But since sin 2 nx + cos 2 nx = 1,

(sin 2 nx + cos 2 nx) dx =

Using (4.5), we get

Then using (4.3) we see that:

The average value (over a period) of sin 2 nx = the average value (over a period) of cos 2 nx

We can say all this more simply in words. By (4.5), the average value of sin 2 nx equals the average value of cos 2 nx. The average value of sin 2 nx + cos 2 nx = 1 is 1. Therefore the average value of sin 2 nx or of cos 2 nx is 1 2 . (In each case the average value is taken over one or more periods.)

PROBLEMS, SECTION 4

1. Show that if f (x) has period p, the average value of f is the same over any interval R a+p of length p. Hint: Write a f (x) dx 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.

R 2. π/2 (a) Prove that sin 2 0 x dx = 0 cos 2 x dx by making the change of variable x= 1 2 π − t in one of the integrals.

R π/2

(b) Use the same method to prove that the averages of sin 2 (nπx/l) and cos 2 (nπx/l) are the same over a period.

In Problems 3 to 12, 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. 3. sin x + 2 sin 2x + 3 sin 3x on (0, 2π)

4. 1−e −x on (0, 1)

350 Fourier Series and Transforms Chapter 7 5. ” cos 2 x “ π

13. Using (4.3) and equations similar to (4.5) to (4.7), show that

sin 2 kx dx = b cos 2 kx dx = 1

(b − a)

if k(b − a) is an integral multiple of π, or if kb and ka are both integral multiples of π/2.

Use the results of Problem 13 to evaluate the following integrals without calculation.