Integration by Parts: Solving Homework Statement

[erok81]

Homework Statement

I had this integral on my physics homework and for the life of me couldn't solve it. I ended up using Maple..well wolframalpha.com because Maple's output sucks.

Anyway here is the problem.

\int_{0}^{\infty} x e^{-2 \alpha x}dx

Homework Equations

\int u dv = uv - \int v du

The Attempt at a Solution

So I set this up as:

u = x
du = dx
dv = e^-2αx
v =

If I was e^x I'd have no problem. Or even integrating only e^-2αx then it'd be a u-sub and no problem.

I tried doing a u-sub right at the beginning and just saved the 1/2α until the end, but that didn't work either.

I am pretty sure once I get to the infinities I'll be able to tackle those. But we can see once we get there.

There is also the tabular method, but I'd rather do integration by parts as I seem to suck at it.

 

[Mark44]

Homework Statement

I had this integral on my physics homework and for the life of me couldn't solve it. I ended up using Maple..well wolframalpha.com because Maple's output sucks.

Anyway here is the problem.

\int_{0}^{\infty} x e^{-2 \alpha x}dx

Homework Equations

\int u dv = uv - \int v du

The Attempt at a Solution

So I set this up as:

u = x
du = dx
dv = e<sup>-2αx[/SUB]
v =

If I was ex I'd have no problem. Or even integrating only e-2αx[/SUB] then it'd be a u-sub and no problem.</sup>

What's the problem? Integrate e-2αx[/SUB] to get v, and you're almost home.

I tried doing a u-sub right at the beginning and just saved the 1/2α until the end, but that didn't work either.

I am pretty sure once I get to the infinities I'll be able to tackle those. But we can see once we get there.

There is also the tabular method, but I'd rather do integration by parts as I seem to suck at it.

I don't much care for it myself.

 

[jhae2.718]

All you need to do to find v is integrate dv, which you seem to indicate you know how to do in your post.

Then put everything into the parts formula, integrate, and apply your limits.

Too late again...

 

[erok81]

So finding v I get

v = \frac{1}{2 \alpha}e^{-2 \alpha x}

Plugging into my integration by parts formula.

<br /> = x \frac{1}{2 \alpha}e^{-2 \alpha x} - \int_{0}^{\infty} \frac{1}{2 \alpha}e^{-2 \alpha x} dx<br />

So if that's right, I can do the same u-sub with u=2α

Which would give me...

\frac{x}{2 \alpha}e^{-2 \alpha x} - \left[ \frac{1}{4 \alpha^{2}}e^{-2 \alpha x} \right] _{0}^{\infty}

Plugging in b for my infinity term, taking the limit as b goes to infinity, then evaluating at 0.

<br /> \int_{0}^{\infty} x e^{-2 \alpha x}dx = \frac{1}{4 \alpha^&amp;{2}}<br />

How's that look? I believe the answer is correct now.

 

[jhae2.718]

So finding v I get

v = \frac{1}{2 \alpha}e^{-2 \alpha x}

You have your sign wrong here. However, you did get the final answer correct, so you must have made an even number of sign errors.

 

[erok81]

I am glad you brought that up. Here is another stupid question.

Mine wasn't negative because my u-sub was 2αx rather than -2αx. I think that's why my answer still right. If I would have u-subbed -2αx I would have ended up with a negative answer.

Checking through the work both ways, I don't think it matters which I choose since the answer still comes out the same. So does it even matter if you include the negative or not?

 

[jhae2.718]

Yes. If you did:
<br /> \int \!\! e^{-2\alpha x} dx<br />
with u=2\alpha x, \quad du = 2\alpha dx, you would end up with:
\int \!\! e^{-2\alpha x} dx = \frac{1}{2\alpha} \int \!\! e^{-u} du = -\frac{1}{2\alpha}e^{-u} = -\frac{1}{2\alpha}e^{-2\alpha x}

 

[erok81]

Got it. Thanks. That's good to know.

It's depressing how quickly one forgets this stuff. :(

Thanks Mark44 for your help as well.