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

Evaluate ∫_(0) ^(1) π‘₯^(2) 𝑒^(π‘₯) dπ‘₯.

02:40

Video Transcript

Evaluate the definite integral between the limits of zero and one of π‘₯ squared times 𝑒 to the power of π‘₯ dπ‘₯.

In this example, we have the product of two functions. That’s a hint that we might need to use integration by parts. This says that the integral of 𝑒 times d𝑣 by dπ‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. So how do we decide what we’re going to let 𝑒 be equal to? Well, remember, we want to ensure that this second integral over here is a little simpler. Therefore, we’re going to want 𝑒 to be a function that either becomes simpler when differentiated or helps to simplify the integrand when multiplied by this 𝑣. It makes no sense for us to let 𝑒 be equal to 𝑒 to the power of π‘₯. Since the derivative of 𝑒 to the power of π‘₯ is just 𝑒 to the power of π‘₯. So, instead, we’re going to let 𝑒 be equal to π‘₯ squared. And d𝑣 by dπ‘₯ is therefore 𝑒 to the power of π‘₯. d𝑒 by dπ‘₯ is then two π‘₯. And the antiderivative of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯. So 𝑣 is 𝑒 to the power of π‘₯. And we obtain the integral to be equal to π‘₯ squared 𝑒 to the power of π‘₯ minus the integral of two π‘₯𝑒 to the power of π‘₯.

Now notice, I’ve used this kind of funny half bracket here. This is just a way of reminding ourselves that we’re dealing with a definite integral. And we’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 𝑒 to the power of π‘₯? Well, we’re going to need to use integration by parts again. For much the same reason, we once again choose d𝑣 by dπ‘₯ to be equal to 𝑒 to the power of π‘₯. And then 𝑒 is equal to two π‘₯. So we see that d𝑒 by dπ‘₯ is equal to two. And 𝑣 is equal to 𝑒 to the power of π‘₯.

So let’s just quickly evaluate the integral of two π‘₯𝑒 to the power of π‘₯. It’s two π‘₯𝑒 to the power of π‘₯ minus the integral of two times 𝑒 to the power of π‘₯. Well, the integral of two 𝑒 to the power of π‘₯ is just two 𝑒 to the power of π‘₯. And I’ve put plus 𝑐 in brackets because the integral we’ve just done is an indefinite integral. But actually the one we’re really going to be doing is between the limits of zero and one. Replacing the integral of two π‘₯𝑒 to the power of π‘₯ with two π‘₯𝑒 to the power of π‘₯ minus two 𝑒 to the π‘₯. And we obtain the integral of π‘₯ squared 𝑒 to the power of π‘₯ to be equal to π‘₯ squared 𝑒 to the power of π‘₯ minus two π‘₯𝑒 to the power of π‘₯ plus two 𝑒 to the power of π‘₯.

Now, we’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 𝑒 to the power of one minus two 𝑒 to the power of one plus two 𝑒 to the power of one. All these disappear and minus two. And we’re done. The integral evaluated between zero and one of π‘₯ squared times 𝑒 to the power of π‘₯ is 𝑒 minus two.

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