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₯.
Weβre 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.
