Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the function $e^ae^{-ba}$

The indefinite integral is $\int e^ae^{-ba} da = \int e^{a-ba} da$ is $\frac{e^ae^{-ba}}{1-b}$

I get that $\int e^u = e^u\frac{du}{da}$.

I cannot seem to understand how the $(1-b)$ term ends up in the denominator.

Can someone point out the rule that I am missing that puts the $(1-b)$ in the denominator?

share|improve this question
    
Briefly, it comes from $e^a e^{-ba} = e^{a-ba} = e^{(1-b)a}$. –  johnny Nov 23 '11 at 17:41
1  
Differentiate $e^{a-ba}$. How does it compare to your integrand? –  David Mitra Nov 23 '11 at 17:43

1 Answer 1

up vote 2 down vote accepted

Note that $a-ba=(1-b)a$. So $$\int e^ae^{-ba}\,da = \int e^{a-ba}\,da = \int e^{(1-b)a}\,da.$$ Setting $u=(1-b)a$, we have $du = (1-b)da$, or $\frac{1}{1-b}\,du = da$. Hence $$\int e^{a}e^{-ba}\,da = \int e^{(1-b)a}\,da = \int \frac{1}{1-b}e^u\,du = \frac{1}{1-b}e^u+C = \frac{e^{(1-b)a}}{1-b}+C.$$

I don't understand what you mean by "$\int e^u = e^u\frac{du}{da}$" (there seem to be lots of things missing, like a $d\Box$ in the integral and a constant of integration).

share|improve this answer
    
There are probably things missing which is probably aiding in my confusion. What I was missing is that $du=(1-b)da; \frac{1}{1-b}du=da$ Thank you. –  strimp099 Nov 23 '11 at 17:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.