### Video Transcript

Evaluate the definite integral
between one and two of π₯ squared times π₯ cubed minus three cubed with respect to
π₯.

This is not a polynomial thatβs
nice to integrate using our standard rules of finding the antiderivative. And we certainly donβt want to
distribute our parentheses and find the antiderivative for each term. Instead, we spot that our integral
is set up in this form. Itβs the definite integral between
some limits of π and π of some function of π of π₯ times the derivative of the
inner part of that composite function. Here, π of π₯, the inner part of
our composite function, is π₯ cubed minus three.

And then, we have a scalar multiple
of its derivative here. So, we use integration by
substitution to evaluate the integral. This says that if π prime, the
derivative of π, is continuous on some closed interval π to π, and π is
continuous on the range of π’, which is our function π of π₯. Then, then the definite integral is
equal to the definite integral between π of π and π of π of π of π’ with
respect to π’. So, we let π’ be equal to the
function that we defined as π of π₯. Itβs the function inside a function
whose derivative also appears.

So, weβre going to let π’ be equal
to π₯ cubed minus three. Now, this is great as when we
differentiate π’ with respect to π₯, we see that dπ’ by dπ₯ equals three π₯
squared. And in integration by substitution,
we think of dπ’ and dπ₯ as differentials. And we can alternatively write this
as dπ’ equals three π₯ squared dπ₯. Notice that whilst dπ’ by dπ₯ is
definitely not a fraction, we do treat it a little like one in this process. We divide both sides by three. And we see that a third dπ’ equals
π₯ squared dπ₯.

So, letβs look back to our original
integral. We now see that we can replace π₯
squared dπ₯ with a third dπ’. And we can replace π₯ cubed minus
three with π’. But what do we do with our limits
of one and two? Well, we need to replace them with
π of one and π of two. Well, we go back to our original
substitution. We said that π’ is equal to π₯
cubed minus three, and our lower limit is when π₯ is equal to one. So, thatβs when π’ is equal to one
cubed minus three, which is equal to negative two.

Our upper limit is when π₯ is equal
to two. So, at this stage, π’ is equal to
two cubed minus three, which is of course five. We can now replace each part of our
integral with the various substitution. And we see that weβre going to need
to work out the definite integral between negative two and five of a third π’ cubed
with respect to π’. Remember, we can take any constant
factors outside of the integral and focus on integrating π’ cubed.

And then, we recall that we can
integrate a polynomial term whose exponent is not equal to negative one by adding
one to that exponent and then dividing by that value. So, the integral of π’ cubed is π’
to the fourth power divided by four. And this means our integral is
equal to a third times π’ to the fourth power divided by four evaluated between
negative two and five. Weβll substitute π’ equals five and
π’ equals negative two, and find their difference.

In this case, thatβs simply a third
of five to the fourth power divided by four minus negative two to the fourth power
divided by four. That gives us a third of 609 over
four. And we can cancel through by
three. And we obtain our solution to be
equal to 203 over four, or 50.75. And so, the definite integral
between one and two of π₯ squared times π₯ cubed minus three cubed with respect to
π₯ is 50.75.

In this example, we saw that we
should try and choose π’ to be some factor of the integrand whose differential also
occurs, albeit some scalar multiple of it. If thatβs not possible though, we
try choosing π’ to be a more complicated part of the integrand. This might be the inner function in
a composite function or similar.