# Video: Finding the Average Value of a Function Involving a Logarithmic Function on a Given Interval Using Integration by Substitution

Find the average value of π(π’) = ln π’/π’ on the interval [1, 3].

02:19

### Video Transcript

Find the average value of π of π’ equals the natural log of π’ over π’ on the closed interval one to three.

We begin by recalling that the average value of a function π over some closed interval π to π is given as one over π minus π times the definite integral evaluated between π and π of π of π₯ with respect to π₯. Now, of course, in this example, weβre dealing with a function in terms of π’. So we redefine the average value as one over π minus π times the definite integral of π of π’ with respect to π’.

Weβre interested in finding the average value of our function over the closed interval one to three. So weβll let π be equal to one and π be equal to three. And this means we can find the average value of our function by calculating one over three minus one or one-half of the definite integral between one and three of the natural log of π’ over π’ with respect to π’. So how do we evaluate that integral?

Well, weβre going to use integration by substitution. We normally define our new function as π’. But since weβre already working with π’, Iβm going to define this new function as π₯. And Iβm going to let π₯ be equal to the natural log of π’. So the derivative of π₯ with respect to π’ is one over π’.

Now, of course, dπ₯ by dπ’ is not a fraction. But when dealing with integration by substitution, we often treat it as one. So we can say that this is equivalent to dπ₯ equals one over π’ dπ’. Weβre also going to redefine our limits by using this substitution. When π’ is equal to one, π₯ is equal to the natural log of one, which is zero. And when π’ is equal to three, π₯ is equal to the natural log of three.

Now look carefully. We replace the natural log of π’ with π₯. And we replace one over π’ dπ’ with dπ₯. And we found that the average value of our function can be now calculated using a half times the definite integral between zero and the natural log of three of π₯ with respect to π₯.

Now the integral of π₯ is just π₯ squared over two. So we need to work out a half of π₯ squared over two evaluated between the natural log of three and zero. Thatβs a half times the natural log of three squared over two minus zero, which we find when we distribute the parentheses to be equal to the natural log of three squared over four. And weβre done. The average value of π of π’ is equal to the natural log of three squared over four.