Integration+by+Parts

=​Integration by Parts - Reuben Luoma-Overstreet=

So... Mr. Clark decides to throw a big end-of-the-year calculus party. He invites all the functions. Everyone is having a blast when he sees his old friend e x sitting in the corner, looking unhappy.
 * Mr. Clark:** "What's wrong e x ?"
 * e x :** "I'm lonely..."
 * Mr. Clark:** "Well then, why don't you integrate yourself in the crowd?"
 * e x :** "It won't make a difference!"


 * Integration by Parts** is a calculus technique that transforms an integral into an integral that is easier to solve.

The simplest form of this rule assumes:

//u// = //f//(//x//) - //du// = //f// '(//x//) //dx// //v// = //g//(//x//) - //dv// = //g//'(//x//) //dx//

and that both are continuously differentiable.

And thus:



Before we continue, try this one on the AP test!

And now, If we take //u// = //f//(//x//) - //u'// =

//v// = //g//(//x//) - //v'// =


 * According to the Product Rule:**




 * And by integrating both sides:**


 * Rearranging this gives us the rule for integration by parts:**

And now, let's see who gets this one: (Don't be a jerk!)

There are two main ways to use integration by parts. Following is an example of each.

This is the simplest form:


 * Integrate** [[image:http://upload.wikimedia.org/math/a/b/3/ab321be463850ef5e2f1d782546bfb1d.png caption="int xcos (x), dx!"]]


 * Solution:**

First, we make a table displaying u and dv
 * **u** || **v'** ||
 * x || cos(x) ||

When we derive u and antiderive dv:

//u// = //x// - //u'// = dx

//v// = //sin(x)// - //v'// = cos(x)

We plug this into the formula to get:

The other technique involves making the original integral reappear and then adding/subtracting it from both sides.


 * Integrate [[image:http://upload.wikimedia.org/math/2/0/e/20ee2b1b32d392fee5b969f623e1a139.png caption="int e^{x} cos (x), dx!"]]**


 * Solution:**

First, we make a table displaying u and dv
 * **u** || **v'** ||
 * cos(x) || e x ||

When we derive u and antiderive dv:

//u// = //cos(x)// - //u'// = -sin(x) //v// = e x - //v'// = e x

We plug this into the formula to get:

Except that this is not the finished product. To finish, we need to perform integration by parts on the last term.

Once more, we make a table displaying u and dv
 * **u** || **v'** ||
 * sin(x) || e x ||

When we derive u and antiderive dv:

//u// = //sin(x)// - //u'// = cos(x) //v// = e x - //v'// = e x

So... the integral...

We then plug this back into the complete equation:

Now, we have made the original term reappear. We use simple algebra to reduce this term out:



And then divide both sides by two:



I apologize for the coding under some of my images - it seems to be unavoidable.

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This is a nice website with many practice problems and solutions: []

And now... Thanks to [] http://www.zazzle.com/calculus_joke_tie-151565750450880691 [] []

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