# Integration by Parts the Adult Way

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• Apr 4th 2008, 12:05 PM
ThePerfectHacker
Integration by Parts the Adult Way
A lot of people do integration by parts by defining the variables $\displaystyle u=...$ and $\displaystyle v'=...$ and flip them around.

There is a more adult way of doing this, which looks nicer and is a lot faster.

Say we have the integral, $\displaystyle \int xe^x dx$
The idea is to turn one of the factors into a derivative. For example, we know that $\displaystyle (e^x)' = e^x$.

Thus, we can think of the integral as,
$\displaystyle \int x \left( e^x \right) ' dx$

The next step is to take the function inside the differenciation operator and multiply it with the function unaffected with the differenciation and multiply them together. That is the $\displaystyle uv$ part that you get.

Thus, we get $\displaystyle xe^x - \int ...$

The next step is to take the derivative of the function which was unaffected by differenciation and multiply it by the function inside the differenciation sign. This is our $\displaystyle u'v$ part.

In this case we get, $\displaystyle xe^x - \int (x) ' e^x dx = xe^x - e^x + C$.
-----
Here is another example,
$\displaystyle \int \ln x dx = \int \ln x \left( x \right) ' dx = x\ln x - \int 1 dx = x\ln x - x + C$.
Look how fast that is.

Here is another example,
$\displaystyle \int x^2 \sin 2x dx = \int x^2 \left( -\frac{1}{2} \cos 2x \right) ' dx = -\frac{1}{2}x^2 \cos 2x + \int x\cos 2x dx$
$\displaystyle =-\frac{1}{2}x^2\cos 2x + \int x \left( \frac{1}{2} \sin 2x \right)' dx = - \frac{1}{2}x^2\cos 2x + \frac{1}{2}x\sin 2x - \int \frac{1}{2}\sin 2x dx =$$\displaystyle -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + C$

My point is that it is a lot easier to keep track of everything doing integration this way. Because you do not need to go out of your way to write $\displaystyle u$ and $\displaystyle v'$.
• Apr 4th 2008, 04:48 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
My point is that it is a lot easier to keep track of everything doing integration this way. Because you do not need to go out of your way to write $\displaystyle u$ and $\displaystyle v'$.

I agree and I think this is quite a clever observation. However, from an educational standpoint, I think teaching this to a student who has never seen integration by parts, ie. Freshman level college, would be a mistake. The students would not likely be able to follow the "reverse product rule" formulation that you worked out.

-Dan
• Apr 4th 2008, 05:00 PM
Krizalid
Quote:

Originally Posted by topsquark
I think teaching this to a student who has never seen integration by parts, ie. Freshman level college, would be a mistake.

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.)
• Apr 5th 2008, 01:12 AM
Gusbob
Quote:

Originally Posted by topsquark
I agree and I think this is quite a clever observation. However, from an educational standpoint, I think teaching this to a student who has never seen integration by parts, ie. Freshman level college, would be a mistake. The students would not likely be able to follow the "reverse product rule" formulation that you worked out.

-Dan

I was taught integration by parts this way... I'm in the 11th grade. What is the other way of doing this?
• Apr 5th 2008, 01:17 AM
janvdl
Quote:

Originally Posted by ThePerfectHacker
Here is another example,
$\displaystyle \int \ln x dx = \int \ln x \left( x \right) ' dx = x\ln x - \int 1 dx = x\ln x - x + C$.
Look how fast that is.

We already do it this way with problems like this. (Nod)

Basically it seems your just skipping the writing of "Let u = ...", and doing that part in your head. I do it often too. (I'm lazy! :D )

EDIT: No not quite skipping it... But I see what you're doing. It is much faster.
• Apr 5th 2008, 02:59 AM
Moo
Well, i agree with topsquark's message, it jumps steps for people who learn it (generally).

Another way to do it is to put f(x) g'(x) dx = f(x) d(g(x)) but it's quite too far for me to remember how the teacher made it...
Some kind of g(x)d(f(x))
• Apr 5th 2008, 05:24 PM
ThePerfectHacker
I also want to add if you have limits of integration you just carry them through. For example,
$\displaystyle \int_0^{\pi} x\cos x dx = \int_0^{\pi} x \left( \sin x \right)' dx = x\sin x \bigg|_0^{\pi} - \int_0^{\pi} \sin x dx = \mbox{ whatever}$.
• Apr 19th 2008, 08:50 AM
Boris B
Dear Perfect H:

I like your method of integration by parts a lot. I haven't tested it yet, so I am still left with one question: is this a complete replacement for u substitution? If so I am very pleased!

Another thing I'm wondering is, are there ever cases in which we can't do the first step, i.e. in which we can't turn one of the factors into a (solvable) derivative? If so, what then?
• Apr 19th 2008, 08:54 AM
colby2152
TPH,

I like the explanation, and like any alternate method or shortcut, it is meant to be learned by the student after they learned the rigorous theorem in the text. Cheers!

-Colby
• Apr 19th 2008, 10:41 AM
Sean12345
Nice, however I'll stick to the original method for my exam (Yes)
• Apr 19th 2008, 01:18 PM
wingless
Quote:

Originally Posted by Boris B
Dear Perfect H:

I like your method of integration by parts a lot. I haven't tested it yet, so I am still left with one question: is this a complete replacement for u substitution? If so I am very pleased!

Another thing I'm wondering is, are there ever cases in which we can't do the first step, i.e. in which we can't turn one of the factors into a (solvable) derivative? If so, what then?

This is another way to do integration by parts, not u substitution (Well, integration by parts includes u substitution). This method is the same as the normal integration by parts, but here we don't have to write u, v, du, dv. If there's an insoluble case, then it means integration by parts can't be used there.

There's also tabular integration (tic-tac-toe method) which makes it easier to work with the cases where you need to apply integration by parts more than once.
• Apr 19th 2008, 08:51 PM
Boris B
I think I've been using this formula wrong. Do we ever actually take the derivative of the thing we originally turned into the derivative? This would be clearer if the example hadn't used a base e exponent; I think that is what may be screwing me up.

I tried to take the antiderivative of:
$\displaystyle f(x) = \frac{2.5(200)^{2.5}}{x^{3.5}}$
First I decided that I could take the derivative of $\displaystyle x^{3.5}$. Since it was in the denominator first, I had to turned into a $\displaystyle x^{-3.5}$ and make it part of the numerator. Then I multiplied that by $\displaystyle 2.5(200)^{2.5}$
$\displaystyle 2.5(200)^{2.5} \cdot x^{-3.5} - \int ...$

For the other side of the integral, I took the derivative of $\displaystyle 2.5(200)^{2.5}$, which, lacking a variable, is 1. That left $\displaystyle x^{3.5}$, which integrates to $\displaystyle \frac{x^{4.5}}{4.5}$

That leaves me with
$\displaystyle 2.5(200)^{2.5} \cdot x^{-3.5} - \frac{x^{4.5}}{4.5} =$
$\displaystyle 1,414,213 x^{-3.5} - \frac{x^{4.5}}{4.5} =$

The definite integrals I take are all quite preposterous (e.g. 5.02 billion minus negative infinity), implying my antiderivative is wrong. (My end goal is to find the difference of the 70th and 30th percentiles of X; I assume I'll need the antiderivative for this but I haven't quite worked out the endgame.)
• Apr 19th 2008, 09:26 PM
Jhevon
I hate using the formula as well. But my way isn't nearly as formal as TPH's. I just always thought of it as: "the integral of one function times the other, minus the $\displaystyle \int$ of the same function times the derivative of the other" sounds confusing in words, but it helps my weird mind to remember. i just remember to integrate one function, and that appears in both factors. then i put the other function in the first factor, and its derivative in the second factor.
• Apr 20th 2008, 04:01 PM
Boris B
To no one in particular
I found the short form of the formula for integration by parts at Wikipedia.

I think I'm finally starting to get it. The short form is:
$\displaystyle \int u dv = uv - \int v du$
Okay ... processing. I think this means that where there is a "u dv" you are actually multiplying variable u by variable dv. Also, dv is the derivative of v. (Why they didn't include a multiplier dot between u and dv is unknown. Edit: it is also unknown why the math script is taking the space out from between u and dv in the above.)

I'm still not sure if I have the correct intepretation here, because in every other case of d_ following an integration symbol the formula did not call for multiplication.

If I'm not mistaken, dx never means multiply, it's just this little thing that follows up integrations or antiderivations (presumably to make the multivariable calculus mavens happy). An example of dx not meaning "multiply by the derivative of x" is two lines up on the Wikipedia page (again "if I'm not mistaken").

I'm really jonesing to get this figured out. Got my fingers crossed.
• Apr 20th 2008, 04:56 PM
Mathstud28
Quote:

Originally Posted by Boris B
I found the short form of the formula for integration by parts at Wikipedia.

I think I'm finally starting to get it. The short form is:
$\displaystyle \int u dv = uv - \int v du$
Okay ... processing. I think this means that where there is a "u dv" you are actually multiplying variable u by variable dv. Also, dv is the derivative of v. (Why they didn't include a multiplier dot between u and dv is unknown. Edit: it is also unknown why the math script is taking the space out from between u and dv in the above.)

I'm still not sure if I have the correct intepretation here, because in every other case of d_ following an integration symbol the formula did not call for multiplication.

If I'm not mistaken, dx never means multiply, it's just this little thing that follows up integrations or antiderivations (presumably to make the multivariable calculus mavens happy). An example of dx not meaning "multiply by the derivative of x" is two lines up on the Wikipedia page (again "if I'm not mistaken").

I'm really jonesing to get this figured out. Got my fingers crossed.

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 (Wink)
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