# Stupid, Easy Anti-Derivative Question.

• Aug 28th 2007, 05:55 PM
SkylerW
Stupid, Easy Anti-Derivative Question.
Sorry I am so stupid... whats the Anti-Derivative of e^x? Again, I apologize for my lack of mathmatical knowledge...
• Aug 28th 2007, 05:57 PM
chocolatelover
Hi,

It's just e^x.

Regards
• Aug 28th 2007, 05:58 PM
SkylerW
So... sorry for being dumb :) Thanks.
• Aug 28th 2007, 06:09 PM
Krizalid
I think is not stupid, it's very interesting.

For example, prove that $\displaystyle (e^x)'=e^x$ or $\displaystyle e=\lim_{x\to\infty}\left(1+\frac1x\right)^x$
• Aug 29th 2007, 06:25 PM
SkylerW
If I could do that... I probably wouldnt need to be taking Calc II :p

Heres another one that is probably very easy for everyone- Antiderivative of ln x
• Aug 29th 2007, 06:26 PM
Krizalid
Use integration by parts.
• Aug 29th 2007, 06:27 PM
SkylerW
Care to walk me through it?... Please :)
• Aug 29th 2007, 06:37 PM
Krizalid
I assume you know the integration by parts formula, so all you need to do it is to set $\displaystyle u=\ln x$ and $\displaystyle dv=dx$, then it's done...
• Aug 29th 2007, 06:41 PM
Jhevon
Quote:

Originally Posted by SkylerW
Care to walk me through it?... Please :)

just as integration by substitution is the integral version of the chain rule (both are used to deal with composite functions), so too is integration by parts similar to the product rule (both are used to deal with the product of functions), in fact, we can kind of derive it from the product rule

recall the product rule for derivatives:

Let $\displaystyle u$ and $\displaystyle v$ be two differentiable functions. By the product rule, we define,

$\displaystyle (uv)' = u'v + uv'$

Now, what if we integrated both sides? we get:

$\displaystyle \int (uv)' = uv = \int u'v + \int uv'$

rearranging this equation, we obtain:

$\displaystyle \boxed { \int uv' = uv - \int u'v }$

this is known as the integration by parts formula. (some books may write $\displaystyle u~dv$ and $\displaystyle v~du$ for $\displaystyle uv'$ and $\displaystyle u'v$ respectively.

Now, we can apply this formula to find the anti-derivative of $\displaystyle \ln x$ in the following way.

since any function can be thought of as itself times one, we think of $\displaystyle \ln x$ here as $\displaystyle 1 \cdot \ln x$. we then let $\displaystyle v' = 1$ and $\displaystyle u = \ln x$ and plug those into the above formula.

You try it, and see what you come up with
• Aug 29th 2007, 09:24 PM
SkylerW
...Maybe I am just not adept enough to understand this kind of thing. Maybe I should drop out of calc II and take calc I. I really just dont understand. Here is my problem, the lnx that I am trying to find the anti-derivative of is the dv already. As in, I am already doing an integration of parts where u=2x and dv=lnx. Therefore du=2 and dv=...? I dont know.

Ill start from the problem in the book. Integrate- (lnx)^2dx. So, I let u=(lnx)^2 and dv=dx. Therefore, du=2lnx and v=x. So, subing it in. I=(lnx)^2 - Integration=2xlnx. Then I let u=2x and dv=lnx. So, du=2 and dv=...?

So thats it. I suppose I am just unintelligent. :)
• Aug 29th 2007, 09:28 PM
Jhevon
Quote:

Originally Posted by SkylerW
...Maybe I am just not adept enough to understand this kind of thing. Maybe I should drop out of calc II and take calc I. I really just dont understand. Here is my problem, the lnx that I am trying to find the anti-derivative of is the dv already. As in, I am already doing an integration of parts where u=2x and dv=lnx. Therefore du=2 and dv=...? I dont know.

Ill start from the problem in the book. Integrate- (lnx)^2dx. So, I let u=(lnx)^2 and dv=dx. Therefore, du=2lnx and v=x. So, subing it in. I=(lnx)^2 - Integration=2xlnx. Then I let u=2x and dv=lnx. So, du=2 and dv=...?

So thats it. I suppose I am just unintelligent. :)

did you get $\displaystyle \int \ln x dx$ ?

if you haven't done the simpler one, why are you jumping to the more complicated one. take it one step at a time, using the pieces Krizalid and i gave you
• Aug 29th 2007, 09:35 PM
SkylerW
I... really just guess I dont understand... I really am sorry.
• Aug 29th 2007, 09:49 PM
Jhevon
Quote:

Originally Posted by SkylerW
I... really just guess I dont understand... I really am sorry.

ok, i will do
$\displaystyle \int \ln x~dx$ for you, then you do $\displaystyle \int ( \ln x )^2~dx$

it is really not as hard as it looks here, we are just pluging things into a formula after we do some calculations

$\displaystyle \int \ln x ~dx$

Let $\displaystyle u = \ln x$ and $\displaystyle \color {blue} v' = 1$

$\displaystyle \Rightarrow \color {red} u' = \frac {1}{x}$ and $\displaystyle \color {green} v = x$

Using $\displaystyle \int u {\color {blue} v'} = u {\color {green} v} - \int {\color {red} u'}{\color {green} v}$

now, match up the colors and/or variables, we get:

$\displaystyle \int \ln x \cdot {\color {blue} 1}~dx = {\color {green} x} \ln x - \int {\color {red} \frac {1}{x}} \cdot {\color {green} x}~dx$

$\displaystyle \Rightarrow \int \ln x~dx = x \ln x - \int ~dx$

...................$\displaystyle = x \ln x - x + C$

ok, so the green font is a little hard to see, but i think you can bear with it