1. Exponential Integration

Our book does a really lousy job of explaining the intricacies of exponential integration. I came upon this problem that I don't know how to solve:

$\int 4^{-2u} du$

I'm sure that there is a chain rule here, so this is what I did:

using the rule $\int a^x dx = \frac {a^x}{\ln a} + C$ I came up with:

$\frac{4^{-u}^2}{\ln 4} + C$

If this doesn't make sense, my exponent comes from the idea that the integral of -2u du is -u^2 (I think).

I know that my answer is wrong, but I don't know what I am doing wrong.

Thanks.

2. Originally Posted by joatmon
Our book does a really lousy job of explaining the intricacies of exponential integration. I came upon this problem that I don't know how to solve:

$\int 4^{-2u} du$

I'm sure that there is a chain rule here, so this is what I did:

using the rule $\int a^x dx = \frac {a^x}{\ln a} + C$ I came up with:

$\frac{4^{-u}^2}{\ln 4} + C$

If this doesn't make sense, my exponent comes from the idea that the integral of -2u du is -u^2 (I think).

Sorry, latex doesn't seem to be working for me, so hopefully this is readable and I have written it correctly.

I know that my answer is wrong, but I don't know what I am doing wrong.

Thanks.
Yes it's wrong. Substitute x = -2u. Then use your formula.

3. But that's what I am asking. How do I handle the -2u in the exponent? Do I just leave it as -2u? Or is there a chain rule or some other process that I need to apply? Since the integral of -2u is -u^2, does that somehow enter in?

Unfortunately, as I mentioned, our book is silent on this, so I really have no idea how to handle this exponent. I wish that I could show more work, but I'm truly stuck.

Thanks again.

4. Originally Posted by joatmon
But that's what I am asking. How do I handle the -2u in the exponent? Do I just leave it as -2u? Or is there a chain rule or some other process that I need to apply? Since the integral of -2u is -u^2, does that somehow enter in?

Unfortunately, as I mentioned, our book is silent on this, so I really have no idea how to handle this exponent. I wish that I could show more work, but I'm truly stuck.

Thanks again.
Haven't you learned how to change an integral using substitution?

Do what I said. The integral becomes $- \frac{1}{2} \int4^x \, dx$. Do it. Then substitute back x = -2u.

Surely you have learned this technique? (If not, then you should not be attempting the posted question until you have).

5. Yes. I do know the technique. We just haven't applied it to a situation where the substitution occurred solely in the exponent. I didn't see that the coefficient 2 would fall out as a constant of 1/2. When you said that x=2u, I didn't realize that you were suggesting a substitution. I just looked at that and saw what I had already done, which was to replace the x in the differentiation rule with the term -2u. Now I see what to do with it. Thanks for your help!

6. BTW, if you didn't know the rule (or if you're like me and only choose to remember a small number of rules), you can convert it to a funtion of e...

.

Which would mean

and you would then make the substitution .