View Single Post
  #1  
Old April 4th, 2008, 02:05 PM
ThePerfectHacker's Avatar
ThePerfectHacker ThePerfectHacker is offline
Global Moderator

 
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,754 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default Integration by Parts the Adult Way

A lot of people do integration by parts by defining the variables u=... and 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, \int xe^x dx
The idea is to turn one of the factors into a derivative. For example, we know that (e^x)' = e^x.

Thus, we can think of the integral as,
\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 uv part that you get.

Thus, we get 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 u'v part.

In this case we get, xe^x - \int (x) ' e^x dx = xe^x - e^x + C.
-----
Here is another example,
\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,
\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
=-\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 =-\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 u and v'.
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
The Following 25 Users Say Thank You to ThePerfectHacker For This Useful Post:
Donate to MHF