Finding the Indefinite Integral of a Product of Exponential Functions

[gyza502]

Homework Statement

F(x)= (3^x)(e^x)dx

Homework Equations

F(u)=U^n=(U^(n+1))/n+1

The Attempt at a Solution

I said it equaled:
((3^(x+1))/(x+1))(e^x)

 

[vela]

Your post doesn't make sense. Why don't you write things out using normal notation so we don't have to guess as to what you mean?

 

[HallsofIvy]

You've made a number of very basic errors. First, and this may be what Vela was complaining about, it doesn't make sense to have a function equal to an integrand- what you meant to say, instead of $F(x)= 3^xe^xdx$ was that the integrand was $3^x e^x dx$ or, equivalently that you were trying to find $F(x)= \int 3^xe^x dx$.

You make the same kind of error when you write "F(u)=U^n=(U^(n+1))/n+1". $U^{n+1}/(n+1)$ is the integral of $U^n$, they are not equal. (Oh, and two minor things- "u" and "U" are not interchangeable and what you wrote, U^(n+1)/n+ 1 is equal to (U^(n+1)/n)+ 1, not U^(n+1)/(n+1).)

Most importantly, that "power rule" does not apply here. It applies to the variable to a constant power and what you have here is a constant to a variable power. And, of course, you cannot simply multiply by $e^x$ as if it were a constant.

Instead, use the fact that $3^x= e^{ln 3^x}= e^{xln(3)}$ and write the integral, $\int 3^xe^x dx$, as $\int e^{x ln(3)}e^x dx= \int e^{x ln(3)+ x}dx= \int e^{x(ln(3)+ 1)}dx$.

Now, do you know how to integrate $\int e^{ax}dx$?

 

[gyza502]

no i do not. can you tell me please?

 

[HallsofIvy]

This isn't for a class? Do you know the derivative of e^ax?

 

[gyza502]

yes it is for a class. But, it is a practice problem. We haven't seen similar problems to this, so i am quite lost at the moment. I know the derivative of e^x is e^x.

 

[SammyS]

yes it is for a class. But, it is a practice problem. We haven't seen similar problems to this, so i am quite lost at the moment. I know the derivative of e^x is e^x.

Do you know the chain rule ?

If so, use it ti find the derivative of e^ax, a being a constant.