### Video Transcript

If the rate of change of a quantity over the closed interval from one to two is given by π prime of π₯ equals π₯ squared times π to the power of two π₯ cubed, what is the net change of the quantity over the closed interval one to two?

To answer this question, weβre going to need to quote the net change theorem. This says that the integral of a rate of change is equal to the net change. That is the definite integral between π and π of capital πΉ prime of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π. Now, weβre told that the rate of change of our quantity over the closed interval from one to two is given by π prime of π₯, which is equal to π₯ squared times π to the power of two π₯ cubed. And it doesnβt matter that weβre dealing with a lowercase π for our function. As long as we ensure that π of π₯ is equal to the antiderivative of π prime of π₯.

And so, to find the net change of our quantity over the closed interval from one to two, weβre going to integrate the rate of change. So, itβs the definite integral between one and two. Those are the bounds of our interval of π₯ squared times π to the power of two π₯ cubed with respect to π₯. Now, this might look like a really complicated integral. But if we spot that π₯ squared is a scalar multiple of the derivative of two π₯ cubed, which is part of a composite function, then that tells us we can use integration by substitution.

Letβs let π’ be equal to two π₯ cubed. Now, we could just let π’ be equal to π₯ cubed, where weβll get the same answer either way. And we choose π’ to be equal to two π₯ cubed. Because when we differentiate π’ with respect to π₯ here, we get six π₯ squared. Remember, we multiply the entire term by the exponent and then reduce the exponent by one. And then, whilst dπ’ by dπ₯ isnβt a fraction, we do treat it a little like one. And we rearrange this expression to get a sixth dπ’ equals π₯ squared dπ₯. And now, we replace π₯ squared dπ₯ with a sixth dπ’. And we replace two π₯ cubed with π’.

But what about the limits? Well, we go back to our earlier substitution, we know π’ was equal to two π₯ cubed. And we know that when π₯ is equal to two, π’, therefore, must be equal to two times two cubed. And that gives us an upper limit then of 16. Then, our lower limit is when π₯ is equal to one. So, we get π’ is equal to two times one cubed, which is equal to two.

And we now have a really nice expression to integrate. The integral of π to the power of π’ is π to the power of π’. And the integral, therefore, of a sixth π to the power of π’ is a sixth π to the power of π’. All thatβs left is to substitute in π’ equals 16 and π’ equals two. And when we do, we find that the net change of our quantity over the closed interval from one to two must be a sixth times π to the power of 16 minus a sixth times π squared.