# MGF Help / General Integration/Multiplication of e^x help please

1. Oct 11, 2008

### Stevieg123

I solved this problem a couple months ago, but seem to have forgotten some rules of calculus with regards to e in the meantime. The goal is to just solve this integral.

Integral from 0 to +inf of (e^tx) times (5e^-5x) dx

Now - in my work I got the answer of 5/5-t, which is correct.

In the key that I have the answer to, one tip they give is that they use the integration rule of:

Integral from 0 to +inf of e^-at dt = 1/a if a>0.

So, I went at it with the standard integration by parts, but quickly realized I had forgotten some rules about multiplying e^ax X e^something else, because I couldn't terminate the integrals to get to the point of reaching 5/5-t.

Just basically did something like:

u= e^tx, dv= 5e^-5x -> du= te^tx, v=: -e^-5x

but I think I forgot a rule somewhere as when I put it into uv - S vdu form I realized I was stuck.

Does anyone know how to solve this? Do I need to do IBP a few times, or can I multiply the e's together to get to the form from before?

-or-

Do I not need to do IBP at all? Is the integral in its original form of

Integral from 0 to +inf (e^tx) times (5e^-5x) dx

able to be multiplied out to get to the e^-at form required in the answer? Looking over it now, I'm starting to figure this may be true. But like I said before, I've forgotten some rules with multiplying the two together. Any help would be appreciated. Thanks.

edit - sorry for the sloppy formatting in some places, I'm unfamiliar with the codes here. I just copied the integral code from someone else's post to make it look sorta nice. The basic integral is (e^tx times 5e^-5x).

2. Oct 11, 2008

### Staff: Mentor

You're making this much harder than it is--you don't need integration by parts. Your original integral is this (where the limits of integration are 0 and $$\infty$$:
$$\int$$ e$$^{tx}$$5e$$^{-5x}$$dx
= 5$$\int e^{(t - 5)x}$$ dx

Here you're integrating something that looks like $$e^{kx}$$, so a simple substitution will suffice.
Also, don't forget that this is an improper integral, so you'll need to take a limit when you evaluate the antiderivative.

3. Oct 11, 2008

### Stevieg123

Thanks, I figured it was something like that. So then my question becomes this. I see how the derivative can easily become this:

5$$\int e^{(t - 5)x}$$ dx

by factoring the 5 out and combining the two.

But my question then becomes, how can it be represented as (t-5) when the answer in front of me clearly states that the integral becomes:

5$$\int e^{-(5-t)x}$$ dx which takes on the form of 5$$\int e^{-at}$$ dt which can then be rewritten as 5/(5-t)? The signs get reversed somewhere in there and it confuses me as to why and how.

Is it just intuition? Am I supposed to look at 5$$\int e^{(t - 5)x}$$ dx and say to myself, oh, just negate the (t-5) and it becomes easy to represent?

4. Oct 12, 2008

### Staff: Mentor

No, you should be using integration, not intuition. You're jumping ahead too many steps without actually doing the intervening work. For this problem it seems that knowing the answer ahead of time is an obstacle.

Here are the things you need to do:
1. Find the antiderivative for the definite integral (limits are 0 and b): 5$$\int e^{(t-5)x}$$ dx. There's actually a limit involved, namely lim b --> $$\infty$$.
2. Evaluate the antiderivative you found at the two limits of integration. You haven't taken the limit yet.
3. In the two expressions you get, one of which involves b, take the limit as b --> $$\infty$$.
If you do these steps correctly you should get the answer you're looking for.

5. Oct 12, 2008

### Staff: Mentor

No, you should be using integration, not intuition. You're jumping ahead too many steps without actually doing the intervening work. For this problem it seems that knowing the answer ahead of time is an obstacle.

Here are the things you need to do:
1. Find the antiderivative for the definite integral (limits are 0 and b): 5$$\int e^{(t-5)x}$$ dx. There's actually a limit involved, namely lim b --> $$\infty$$.
2. Evaluate the antiderivative you found at the two limits of integration. You haven't taken the limit yet.
3. In the two expressions you get, one of which involves b, take the limit as b --> $$\infty$$.
If you do these steps correctly you should get the answer you're looking for.