Video Transcript
Evaluate the definite integral between
the limits of zero and one of π₯ squared times π to the power of π₯ dπ₯.
In this example, we have the product of
two functions. Thatβs a hint that we might need to use
integration by parts. This says that the integral of π’ times
dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯. So how do we decide what weβre going to
let π’ be equal to? Well, remember, we want to ensure that
this second integral over here is a little simpler. Therefore, weβre going to want π’ to be
a function that either becomes simpler when differentiated or helps to simplify the
integrand when multiplied by this π£. It makes no sense for us to let π’ be
equal to π to the power of π₯. Since the derivative of π to the power
of π₯ is just π to the power of π₯. So, instead, weβre going to let π’ be
equal to π₯ squared. And dπ£ by dπ₯ is therefore π to the
power of π₯. dπ’ by dπ₯ is then two π₯. And the antiderivative of π to the
power of π₯ is π to the power of π₯. So π£ is π to the power of π₯. And we obtain the integral to be equal
to π₯ squared π to the power of π₯ minus the integral of two π₯π to the power of π₯.
Now notice, Iβve used this kind of
funny half bracket here. This is just a way of reminding
ourselves that weβre dealing with a definite integral. And weβre going to need to evaluate
both parts of it between the limits of zero and one. But how do we evaluate the integral of
two π₯ times π to the power of π₯? Well, weβre going to need to use
integration by parts again. For much the same reason, we once again
choose dπ£ by dπ₯ to be equal to π to the power of π₯. And then π’ is equal to two π₯. So we see that dπ’ by dπ₯ is equal to
two. And π£ is equal to π to the power of
π₯.
So letβs just quickly evaluate the
integral of two π₯π to the power of π₯. Itβs two π₯π to the power of π₯ minus
the integral of two times π to the power of π₯. Well, the integral of two π to the
power of π₯ is just two π to the power of π₯. And Iβve put plus π in brackets
because the integral weβve just done is an indefinite integral. But actually the one weβre really going
to be doing is between the limits of zero and one. Replacing the integral of two π₯π to
the power of π₯ with two π₯π to the power of π₯ minus two π to the π₯. And we obtain the integral of π₯
squared π to the power of π₯ to be equal to π₯ squared π to the power of π₯ minus two
π₯π to the power of π₯ plus two π to the power of π₯.
Now, weβre going to need to evaluate
this between the limits of zero and one. Substituting zero and one and finding
their difference. And we have π to the power of one
minus two π to the power of one plus two π to the power of one. All these disappear and minus two. And weβre done. The integral evaluated between zero and
one of π₯ squared times π to the power of π₯ is π minus two.
