Uploaded by TheIntegralCALC on 19.07.2010

Transcript:

Hi everyone! Welcome back to integralcalc.com. We're going to be doing another integration

by parts problem today. This one is the integral of x times e to the x dx.

And the reason that we know that this is an integration by parts problem is because we

have two components here in the integral that comprise the entire integral. It's not a partial

fractions problem because we don't have a fraction here. It could be u-substitution

but you're always looking for something to cancel when you use u-substitution and you're

always looking for the derivative of u to be something that will cancel out something

that's already in your integral. So for example, if we assign u to e to the x, the derivative

is e to the x. That's not going to cancel out the x here. And if we assign u to x, the

derivative of x is just one and that's certainly not going to cancel anything out.

So let's try integration by parts. Remember the integration by parts equation is the integral

of u times dv equals u v minus the integral of v du.

So this is the equation we're working with and we need to assign u and dv to identities

in our integral here. So what we, the easiest way to do it is to assign u to something that

will become simpler when you take its derivative. So in this case, that's definitely x. The

derivative of e to the x is still e to the x which does not simplify at all but the derivative

of x is one, which is much simpler than x. So we're going to assign u to x, x and then

we need to assign dv, by default to the other value in our problem, e to the x. And from

u and dv, we need to find du and v. We'll find du by taking the integral, the derivative

of u and the derivative of x is just one. v, we determine by taking the integral of

e to the x. The integral of e to the x is still e to the x.

So now that we've found these four identities, we can go ahead and plug them into the second

part or the right side of this equation here. So you can see, we have u times v. So here,

u times v. So, x times e to the x minus the integral of v, which we have is e to the x

times du, so times one and then we put in the dx here as notation. This one's going

to go away, (right?) because it's just redundant. So we're just left with the integral of e

to the x, which is e to the x. So x times e to the x minus, we take the integral and

we just flip e to the x and then we add c to account for the constant. And this is actually

our final answer. If you would like to, you can factor out an

e to the x, so you either have x e to the x minus e to the x plus c or if you want to

factor an e to the x, you can do so. I'm not sure which way is simpler. You'd have x

minus one plus c. Either way, they're equally correct but that's

your final answer. Hope that helped. See you guys next time! Bye!

by parts problem today. This one is the integral of x times e to the x dx.

And the reason that we know that this is an integration by parts problem is because we

have two components here in the integral that comprise the entire integral. It's not a partial

fractions problem because we don't have a fraction here. It could be u-substitution

but you're always looking for something to cancel when you use u-substitution and you're

always looking for the derivative of u to be something that will cancel out something

that's already in your integral. So for example, if we assign u to e to the x, the derivative

is e to the x. That's not going to cancel out the x here. And if we assign u to x, the

derivative of x is just one and that's certainly not going to cancel anything out.

So let's try integration by parts. Remember the integration by parts equation is the integral

of u times dv equals u v minus the integral of v du.

So this is the equation we're working with and we need to assign u and dv to identities

in our integral here. So what we, the easiest way to do it is to assign u to something that

will become simpler when you take its derivative. So in this case, that's definitely x. The

derivative of e to the x is still e to the x which does not simplify at all but the derivative

of x is one, which is much simpler than x. So we're going to assign u to x, x and then

we need to assign dv, by default to the other value in our problem, e to the x. And from

u and dv, we need to find du and v. We'll find du by taking the integral, the derivative

of u and the derivative of x is just one. v, we determine by taking the integral of

e to the x. The integral of e to the x is still e to the x.

So now that we've found these four identities, we can go ahead and plug them into the second

part or the right side of this equation here. So you can see, we have u times v. So here,

u times v. So, x times e to the x minus the integral of v, which we have is e to the x

times du, so times one and then we put in the dx here as notation. This one's going

to go away, (right?) because it's just redundant. So we're just left with the integral of e

to the x, which is e to the x. So x times e to the x minus, we take the integral and

we just flip e to the x and then we add c to account for the constant. And this is actually

our final answer. If you would like to, you can factor out an

e to the x, so you either have x e to the x minus e to the x plus c or if you want to

factor an e to the x, you can do so. I'm not sure which way is simpler. You'd have x

minus one plus c. Either way, they're equally correct but that's

your final answer. Hope that helped. See you guys next time! Bye!