1. ## integrate

Show that the integral of e^(-kx)dx from 0 to 1 is 1/k(1-e^-k).
When I integrate from and calculate from 0 to 1, I get k(1-e^-k) instead of 1/k.

My solution is...

[-ke^(-kx)] upper limit 1, lower limit 0
= -ke^(-k) +k
= k(-e^(-k) +1)

Where did I go wrong?

2. $\int e^{-kx}~dx = \frac{-1}{k}e^{-kx} \neq -ke^{-kx}$

3. I don't follow...

4. Originally Posted by bhuang
I don't follow...
let $u = -kx \implies du = -k dx$

$\therefore dx = \frac{du}{-k}$

$\int \frac{e^u \times du}{-k} = \frac{-1}{k} e^u$

substitute back u = -kx, you have:

$\frac{-1}{k} e^{-kx}$. Plug in the upper and lower limits in this expression.

5. Thanks! I can do it through substitution, but I'm just wondering why the way I did it did not work. I just followed the rule that the derivative of a function in the form: f(x) = e^(g(x)) is e^(g(x)) x g'(x).

6. Originally Posted by bhuang
Thanks! I can do it through substitution, but I'm just wondering why the way I did it did not work. I just followed the rule that the derivative of a function in the form: f(x) = e^(g(x)) is e^(g(x)) x g'(x).
Because that is the rule for a derivative, you are integrating.

7. woooooooooow :| thanks!