# Video: Finding the Integral of a Function Using Integration by Parts

Evaluate β«_(0) ^(1) π₯^(2) π^(π₯) dπ₯.

02:40

### 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. Thatβs 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 weβre going to let π’ be equal to? Well, remember, we want to ensure that this second integral over here is a little simpler. Therefore, weβre 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, weβre 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, Iβve used this kind of funny half bracket here. This is just a way of reminding ourselves that weβre dealing with a definite integral. And weβre 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, weβre 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 letβs just quickly evaluate the integral of two π₯π to the power of π₯. Itβs 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 Iβve put plus π in brackets because the integral weβve just done is an indefinite integral. But actually the one weβre 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, weβre 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 weβre done. The integral evaluated between zero and one of π₯ squared times π to the power of π₯ is π minus two.