# integrate

• April 21st 2010, 04:30 PM
bhuang
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?
• April 21st 2010, 04:36 PM
pickslides
$\int e^{-kx}~dx = \frac{-1}{k}e^{-kx} \neq -ke^{-kx}$
• April 21st 2010, 04:54 PM
bhuang
I don't follow...
• April 21st 2010, 05:00 PM
harish21
Quote:

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.
• April 21st 2010, 05:08 PM
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).
• April 21st 2010, 05:20 PM
pickslides
Quote:

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.
• April 21st 2010, 05:29 PM
bhuang
woooooooooow :| thanks!