Author Topic: integration (Read 958 times) Tweet Share

jasoN- · « **on:** March 28, 2011, 09:58:08 pm »

how do u integrate $a^{-x}$

from wolfram:

idk how they got this, cos i just assumed a is similar to e (as in e = 2.7......)
so $\int a^{-x}dx=-a^{-x} +c$
it's probably something very simple, i know! but i just haven't learnt it.

any help appreciated

TrueTears · « **Reply #1 on:** March 28, 2011, 10:03:36 pm »

jasoN- · « **Reply #2 on:** March 28, 2011, 10:16:57 pm »

sorry im still confused, can i get the full working out? lolz

thanks

jasoN- · « **Reply #3 on:** March 28, 2011, 11:02:40 pm »

nvm i understand it now:

$a^{-x} = e^{log_e{a^{-x}}}$

$a^{-x} = e^{-xlog_e{a}}$

$\frac{d(a^{-x})}{dx} = -log_e{a}*a^{-x}$

hence
$d(a^{-x}) = (-log_e{a}*a^{-x}) dx$

$\int d(a^{-x}) = -log_e{a} \int(a^{-x}) dx$

$\int(a^-x) dx = -\frac{a^{-x}}{log_e{a}} + c$

does $\int d(a^{-x}) = a^{-x}$ make sense?

Mao · « **Reply #4 on:** March 30, 2011, 04:11:51 am »

Quote from: jasoN- on March 28, 2011, 11:02:40 pm

does $\int d(a^{-x}) = a^{-x}$ make sense?

With a change of variables, the argument is the same as $\int\, du = u + C$ for any $u$ . That statement is fine.

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