Video Transcript
Evaluate the definite integral between
the limits of zero and one of 𝑥 squared times 𝑒 to the power of 𝑥 d𝑥.
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 𝑢 times
d𝑣 by d𝑥 is equal to 𝑢𝑣 minus the integral of 𝑣 times d𝑢 by d𝑥. So how do we decide what we’re going to
let 𝑢 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 𝑢 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 𝑢 be
equal to 𝑒 to the power of 𝑥. Since the derivative of 𝑒 to the power
of 𝑥 is just 𝑒 to the power of 𝑥. So, instead, we’re going to let 𝑢 be
equal to 𝑥 squared. And d𝑣 by d𝑥 is therefore 𝑒 to the
power of 𝑥. d𝑢 by d𝑥 is then two 𝑥. And the antiderivative of 𝑒 to the
power of 𝑥 is 𝑒 to the power of 𝑥. So 𝑣 is 𝑒 to the power of 𝑥. And we obtain the integral to be equal
to 𝑥 squared 𝑒 to the power of 𝑥 minus the integral of two 𝑥𝑒 to the power of 𝑥.
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 𝑥 times 𝑒 to the power of 𝑥? Well, we’re going to need to use
integration by parts again. For much the same reason, we once again
choose d𝑣 by d𝑥 to be equal to 𝑒 to the power of 𝑥. And then 𝑢 is equal to two 𝑥. So we see that d𝑢 by d𝑥 is equal to
two. And 𝑣 is equal to 𝑒 to the power of
𝑥.
So let’s just quickly evaluate the
integral of two 𝑥𝑒 to the power of 𝑥. It’s two 𝑥𝑒 to the power of 𝑥 minus
the integral of two times 𝑒 to the power of 𝑥. Well, the integral of two 𝑒 to the
power of 𝑥 is just two 𝑒 to the power of 𝑥. 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 𝑥𝑒 to
the power of 𝑥 with two 𝑥𝑒 to the power of 𝑥 minus two 𝑒 to the 𝑥. And we obtain the integral of 𝑥
squared 𝑒 to the power of 𝑥 to be equal to 𝑥 squared 𝑒 to the power of 𝑥 minus two
𝑥𝑒 to the power of 𝑥 plus two 𝑒 to the power of 𝑥.
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 𝑥 squared times 𝑒 to the power of 𝑥 is 𝑒 minus two.
