Integration by Parts the Adult Way - Page 2

Boris B · #16 $Old$ April 20th, 2008, 07:23 PM

"You are almost right except u and v are functions of x...if they were different variables you would have to assume one is a constant...making the integration exceedingly simple"

Okay, that may be where I'm going wrong. When I have taken u and v in the past, one of them is almost always a constant, yielding a derivative of zero. While the integration didn't seem exceedingly simple, the answers have usually been wrong.

Back to the drawing board.

Mathstud28 · #17 $Old$ April 20th, 2008, 07:25 PM

Quote:

Originally Posted by Boris B $View Post$

"You are almost right except u and v are functions of x...if they were different variables you would have to assume one is a constant...making the integration exceedingly simple"

Okay, that may be where I'm going wrong. When I have taken u and v in the past, one of them is almost always a constant, yielding a derivative of zero. While the integration didn't seem exceedingly simple, the answers have usually been wrong.

Back to the drawing board.

Here try looking at this

Derivative Tutorial

it should explain

Mathstud28 · #18 $Old$ April 20th, 2008, 07:33 PM

Quote:

Originally Posted by Boris B $View Post$

"You are almost right except u and v are functions of x...if they were different variables you would have to assume one is a constant...making the integration exceedingly simple"

Okay, that may be where I'm going wrong. When I have taken u and v in the past, one of them is almost always a constant, yielding a derivative of zero. While the integration didn't seem exceedingly simple, the answers have usually been wrong.

Back to the drawing board.

and here is an example $\int{x\cdot{e^{x}}dx}$ letting $u=x$ and $dv=e^{x}dx$ we see that $du=dx$ and $v=e^{x}$ ...now using the formula you mentioned set this up

Boris B · #19 $Old$ April 21st, 2008, 11:37 PM

Please allow me to rephrase some rules about integration by parts, so you can correct them if I have it wrong.

1. Integration by parts is necessary if you are integrating a formula with anything in other than constants (known or unknown) and a single appearance of one variable. Thus:
$\int ax^2 dx$
must be integrated by parts if a and x are variables; it need not be if a is a constant (I haven't done this with multivariable calculus; I'm just implying here that a is some expression with x in it). Constants can be moved to the left of the integral as always.

2. When looking at the parts of a function, something only counts as a "part" if it is a factor. Addends don't count unless they can be put into factor form. So $x^2 + x$ can't be broken into parts until you turn it into $x \cdot (x+1)$

3. Before you integrate by parts you'll need to find a factor whose derivative is also in the equation, and it's perfectly find to use one whose factor is one or itself. I.e. if dv = v or dv = 1 you are still good to go.

That's my current understanding of this whole thing. Some of the rules were kind of subtle, probably because they were aimed at people who haven't gone 13 years without a math class.

Boris B · #20 $Old$ April 23rd, 2008, 11:48 PM

I started with distribution function
$F(x) = 0.5 (1 + sin \pi x)$ for $1.5 \leq x \leq 2.5$

I derived it to
$f(x) = 0.5 cos \pi x$

Then I'm supposed to get the expected value by multiplying Y into that formula and integrating
$E(x) = \int \pi x \cdot 0.5 cos \pi x dx$
I set $v = \pi x$
$dv = \pi$
$u = 0.5 cos \pi x^2$
$du = -0.5 sin \pi x^2$

I was a little uncertain about it, because in most examples I've seen, v = dv.

It didn't really get me anywhere
$E(x) = \int \pi x \cdot 0.5 cos \pi x dx$
$E(x) = \int 0.5 cos \pi x^2 \cdot \pi$
$E(x) = 0.5 cos \pi^2 x^3 - \int \pi x \cdot sin \pi x$

I suppose one of the reasons I can never get integration by parts (or any integral calculus beyond the power rule) to work is that I always end up with more complexity than I started with. I assume $\int \pi x \cdot sin \pi x$ will have to be solved by integration by parts, is that correct?

Isomorphism · #21 $Old$ April 24th, 2008, 12:10 AM

Quote:

Originally Posted by Boris B $View Post$

I\snip\

Then I'm supposed to get the expected value by multiplying Y into that formula and integrating
$E(x) = \int \pi x \cdot 0.5 cos \pi x dx$
I set $v = \pi x$
$dv = \pi$
$u = 0.5 cos pi x^2$
$du = -0.5 sin \pi x^2$

\snip\

Why did the x^2 suddenly jump in?
So your integral is
$E(x) = \frac1{2\pi} \int u \cdot \cos u du$

Integration by parts is $\int u \, dv = vu - \int v \, du$ .

Now if $\int ab \, dx$ is to be evaluated, where a and b are functions of x, we need to choose between $du = b \, dx$ or $du = a \, dx$ .
This means we have to differentiate a in the first choice and integrate $b \, dx$ to get the right side of the equation for Integration by parts. The $\int \left(b \, dx\right) \, da$ could need integration by parts again. See you want to stop integrations as fast as possible,right? Which will stop faster? If you keep on differentiating 'x' or 'sin x'? Think about it.

Now lets work this out. Differentiating u successively stops in 1 differentiation.

So choose $u = u$ and $dv = \cos u \, du$

So that $v = \sin u$ and $du = du$

Now $\int u \, dv = vu - \int v \, du$ , so:

$\int u \cos u \, du = u\sin u - \int \sin u \, du = u\sin u + \cos u + C$

Boris B · #22 $Old$ April 24th, 2008, 04:39 PM

It sounds like u is usually going to be x so du is 1, because 1 and usually x are in the expression to be integrated. I guess that's the part I'm having trouble getting ... integrating x seems like such a trivial process that I often skip over it in favor of something else.

Mathstud28 · #23 $Old$ April 24th, 2008, 04:42 PM

Quote:

Originally Posted by Boris B $View Post$

It sounds like u is usually going to be x so du is 1, because 1 and usually x are in the expression to be integrated. I guess that's the part I'm having trouble getting ... integrating x seems like such a trivial process that I often skip over it in favor of something else.

Yes $x\cdot{f(x)}$ is usually a preferable usage for integration by parts since it yields $\int{x\cdot{f(x)}dx}=x\int{f(x)}dx-\int\int{f(x)}dxdx$ which is usually easy to do as in thie case $\int{x\cdot{e^{x}}}dx$

Nayomi · #24 $Old$ August 19th, 2008, 02:30 AM

It seems to be a good method. but I can not imagine whether my master would agree with this method.

TKHunny · #25 $Old$ August 19th, 2008, 05:44 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

A lot of people do integration by parts by defining the variables $u=...$ and $v'=...$ and flip them around.

Wow! How did I miss out on this conversation?!

When I was first introduced to Integration by Parts the square table way, by my American Professor, u = blah, dv = blah, I did not like it. I found it confusing. My major objection was that is was only an indirect application of the simplest form of the concept.

$\int u\; dv = uv - \int v\; du$

There is no square table in there. I resisted creating it.

After a two year absence off doing missionary work, I took this course over again, just as a refresher. My Russian professor opened my eyes to what I call the DIRECT applciation of the rule. I would take it one step father than has been done in the example that started this thread. It makes it very often more visually efficient.

$\int ln(x)\; dx = xln(x) - \int x\; d(ln(x)) = xln(x) - \int\; dx$

I realize something besides 'x' hanging around with the 'd' may be a bit shocking.

$\int xe^{x}\; dx = \int x\; d(e^{x}) = xe^{x} - \int e^{x}\; dx$

It's faster. It's less confusing. It's more visual. It's a DIRECT application of the rule. No need to put down the rule to make a table and then put the rule back together.

$\int xln(x) dx = \int ln(x) d\left(\frac{1}{2}x^{2}\right) = \frac{1}{2}x^{2}ln(x) - \int \frac{1}{2}x^{2} d(ln(x)) = \frac{1}{2}x^{2}ln(x) - \int \frac{1}{2}x dx$

It is perfectly valid. It does not constitute a "short cut" that really isn't as rigorious as the "real" way. If you do this on an exam and lose marks, please call me. I will be happy to argue it with whoever doesn't get it.

My views. I welcome others'.

ThePerfectHacker · #26 $Old$ August 19th, 2008, 11:26 AM

Quote:

Originally Posted by Nayomi $View Post$

It seems to be a good method. but I can not imagine whether my master would agree with this method.

Tell your professor to go kill himself if he says something like that.
It does not matter how you integrate as long as you know what you are doing.

Black Kawairothlite · #27 $Old$ November 4th, 2008, 08:06 PM

thanks i like this way seems to be easier,

i always hated to write out the substitutions hehe

Quote:

Originally Posted by Krizalid $View Post$

That's pretty obvious. This method is mechanical, it's for people who have covered integration by parts. (I'm not sayin' the method doesn't work, I'm answering to Dan's post.)

The method is cool and I recently translated to post it in my spanish forum. (Not mine, of course.)

what spanish forum?! link me hehe

Krizalid · #28 $Old$ November 5th, 2008, 03:01 AM

Portal Matemático.

tom@ballooncalculus · #29 $Old$ November 5th, 2008, 12:49 PM

Hey, I'm thrilled to discover this thread! Grateful to get any of your views on a diagramming method that I intended just for doing integration by parts by visualising the product rule - and then elaborated for the chain rule as you'll see elsewhere on the page.

Would very much hope this might please those people here who wanted a method to offer beginning calculus students, or who wanted something 'visual', or who yearned to by-pass 'u substitution' altogether. I'm also fascinated by some of the discussion here that touches particularly on this thread - after which I feel rather less guilty than I did about employing such a reckless slogan as 'say goodbye to dx/dy!' on my homepage.

Will post some pics for this thread (if that's ok), but have to dash now - please click the above, or google. Cheers
Tom

devinhug4free · #30 $Old$ March 23rd, 2009, 11:28 PM

Indeed very nice an even easier method is to set up a chart (I believe this was what other people were reffering to when talking of a tabular form?). For example $\int x^2sinxdx$

$\begin{array}{c|c|c|c}sign&u&dv&Result\\\hline&&sinx&\\\hline+&x^2&-cosx&-x^2cosx\\\hline-&2x&-sinx&2xsinx\\\hline+&2&cosx&2cosx\end{array}$ Answer $-x^2cosx+2xsinx +2cosx+C$

The sign column is just the alternating signs as far down as you need to go. The u column is found by placing u in the third row of it and taking the derivative of each succesive place untill you reach a constant. the dv column is ofund by placing the dv in the second row and taking the succesive integrals of each answer. The results column is found by multiplying the first three rows for each column that has a u value. The answer to the derivative is the sum of the results column. This technique only works when the u term will go to a constant after susessive derivatives. You can fudge it to make it work at other times but it can be tricky. But if you have something that is realy straight forward but will take a number of repeated integrations by parts.
Another example $\int (x^3+2x^2+3)e^xdx$

$\begin{array}{c|c|c|c}sign&u&dv&Result\\\hline&&e^x&\\\hline+&x^3+2x^2+3&e^x&x^3e^x+2x^2e^x+3e^x\\\hline-&3x^2+4x&e^x&-3x^2e^x-4xe^x\\\hline+&6x+4&e^x&6xe^x+4e^x\\\hline-&6&e^x&-6e^x\end{array}$ Answer $x^3e^x-x^2e^x+2xe^x+e^x$