Question Video: Finding the Average Value of a Function on a Given Interval Involving Using Integration by Substitution | Nagwa Question Video: Finding the Average Value of a Function on a Given Interval Involving Using Integration by Substitution | Nagwa

# Question Video: Finding the Average Value of a Function on a Given Interval Involving Using Integration by Substitution Mathematics • Higher Education

Find the average value of π(π₯) = π₯Β²/(π₯Β³ β 5)Β² on the interval [β1, 1].

02:42

### Video Transcript

Find the average value of π of π₯ equals π₯ squared over π₯ cubed minus five all squared on the closed interval negative one to one.

Remember, the formula for the average value of a function over a closed interval π to π is given as one over π minus π times the integral of π of π₯ with respect to π₯ evaluated between π and π. In this case, we can see that our function is equal to π₯ squared over π₯ cubed minus five all squared. And we see our closed interval is from negative one to one. So we let π be equal to negative one and π be equal to one. And this means the average value of our function is given as one over one minus negative one multiplied by the integral of π₯ squared over π₯ cubed minus five all squared with respect to π₯ evaluated between negative one and one.

One minus negative one is two. But how do we evaluate this definite integral? Well, we need to notice that the numerator is a scalar multiple of the derivative of part of the denominator. The derivative of π₯ cubed minus five is three π₯ squared. And this tells us we can use integration by substitution. We let π’ be equal to π₯ cubed minus five, and dπ’ by dπ₯ is therefore equal to three π₯ squared. We know that dπ’ by dπ₯ is not a fraction. But for the purposes of integration by substitution, we do treat it a little like where we say that this is equivalent to a third dπ’ equals π₯ squared dπ₯.

Weβre now in a position to replace the various parts of our integral. We replace π₯ squared dπ₯ with a third dπ’. And we replace π₯ cubed minus five with π’. And we see that the average value of our function is equal to a half times the integral of a third times one over π’ squared with respect to π’. But what do we do about these limits?

Well, we use our definition of π’. We said that π’ was equal to π₯ cubed minus five. So for our upper limit, when π₯ is equal to one, π’ is equal to one cubed minus five, which is negative four. And when π₯ is equal to negative one, π’ is equal to negative one cubed minus five, which is equal to negative six. We can take the constant a third outside of the integral sign and rewrite one over π’ squared as π’ to the negative two. And we know that the integral of π’ to the negative two is π’ to the negative one divided by negative one, or negative π’ to the negative one, which can then be written as negative one over π’.

We substitute negative four and negative six. And we get a sixth of negative one over negative four minus negative one over negative six. Well, negative one over negative four is just a quarter. And negative one over negative six is a sixth. We subtract these fractions by creating a common denominator. And we see that the average value of our function is a sixth times a twelfth, which is one over 72.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions