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.
