Stupid, Easy Anti-Derivative Question.

**SkylerW** · August 28th 2007, 05:55 PM

Sorry I am so stupid... whats the Anti-Derivative of e^x? Again, I apologize for my lack of mathmatical knowledge...

**chocolatelover** · August 28th 2007, 05:57 PM

Hi,

It's just e^x.

Regards

**SkylerW** · August 28th 2007, 05:58 PM

So... sorry for being dumb

Thanks.

**Krizalid** · August 28th 2007, 06:09 PM

I think is not stupid, it's very interesting.

For example, prove that $(e^x)'=e^x$ or $e=\lim_{x\to\infty}\left(1+\frac1x\right)^x$

**SkylerW** · August 29th 2007, 06:25 PM

If I could do that... I probably wouldnt need to be taking Calc II

Heres another one that is probably very easy for everyone- Antiderivative of ln x

**Krizalid** · August 29th 2007, 06:26 PM

Use integration by parts.

**SkylerW** · August 29th 2007, 06:27 PM

Care to walk me through it?... Please

**Krizalid** · August 29th 2007, 06:37 PM

I assume you know the integration by parts formula, so all you need to do it is to set $u=\ln x$ and $dv=dx$ , then it's done...

**Jhevon** · August 29th 2007, 06:41 PM

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

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

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

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

rearranging this equation, we obtain:

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

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

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

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

You try it, and see what you come up with

**SkylerW** · August 29th 2007, 09:24 PM

...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.

**Jhevon** · August 29th 2007, 09:28 PM

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 $\int \ln x dx$ ?

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

**SkylerW** · August 29th 2007, 09:35 PM

I... really just guess I dont understand... I really am sorry.

**Jhevon** · August 29th 2007, 09:49 PM

Originally Posted by SkylerW

I... really just guess I dont understand... I really am sorry.

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

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

$\int \ln x ~dx$

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

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

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

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

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

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

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

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

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