# Thread: Another integral, general power formula.

1. ## Another integral, general power formula.

Hi thanks for the help lately its been good. I don't post these unless a day goes by without a solution.

28-1-20

Integrate:

(int) 2 sqrt(1- e^(-x)) (e^-x) dx

therefore,
u = 1 - e^(-x)
du = - e^(-x)

General power form:
[1 - e^(-x)]^(3/2) / (3/2) + C
=
4/3 [1 - e^(-x)]^(3/2) + C

however here is the ti-89 answer:

somehow something has been added in the nominator.
how did I go wrong? Thanks!

2. ## Re: Another integral, general power formula.

Hello, togo!

$\int \sqrt{1-e^{-x}}\cdot e^{-x}\,dx$

Therefore:

$u \:=\: 1 - e^{-x}$

$du \:=\: - e^{-x}dx$ . . . . no

$\text{If }u \:=\:1-e^{-x}$

. . $\text{then: }\:du \:=\:\left[0-e^{-x}(\text{-}1)\right]dx \:=\:e^{-x}dx$

3. ## Re: Another integral, general power formula.

but, that doesn't really change what I did at all

its the added e^(-3x/2) that is confusing me in the nominator

4. ## Re: Another integral, general power formula.

Originally Posted by togo
but, that doesn't really change what I did at all

its the added e^(-3x/2) that is confusing me in the nominator
Well it DOES change what you did, because if you followed your own logic, you should have ended up with a negative answer. I'm not sure where your negative has gone.

5. ## Re: Another integral, general power formula.

I have been having trouble with losing the negative and have retraced it twice in homework the last couple days.

This is something I can do myself and I'm not sure what to do in order to complete the answer besides this. I'll rework the question soon and hopefully it does give me the answer shown on the ti-89.

6. ## Re: Another integral, general power formula.

ya that change does not give me a complete answer.