Tuesday, February 20, 2007

7.1 - Integration by Parts


Alright, folks, the Lord Protector of Math is back for some more Calculus action! Huzzah!

This lesson will be about integration by parts. This is somewhat like the substitution rule, but it deals with the Product Rule instead of the Chain Rule.

The basics of integration by parts are pretty easy. Let's start with the basic formula.




If we were to use u and v in place of the functions, the formula would simplify to this.




This is taken from the product rule for derivatives, and it allows us to integrate when two functions are multiplied.

Let's get an example. Evaluate the integral.




First, we go about setting values. There will be u, v, du and dv. It's generally a good idea to set the simplest function as the u value and the more complicated function as the dv value.
So, let's set x as u, and sin2x as dv.
This gives us:Since dx is the derivative of x and -(1/2)cos2x is the antiderivative of sin2xdx.
Are you with me? We're almost done.

Now we plug these equations into the equation.



See? We just plugged in the equations for u, du, v, and dv.
Now we evaluate the indefinite integral.


Don't forget your + C, or I will laugh at you pretentiously.
Anyway, see where we're going with this? Let's finish up.


Done! All evaluated.

There are some other things you need to remember:

1.You may have to repeat the process more than once to get rid of all of the integrals.
2. If the functions keep repeating, continue until you have reached an equation similar to the starting problem. Then, add/subtract it to one side, and solve.

Finally, the ultimate time-saver!

THE TABULAR METHOD

This is pretty straightforward and easy. If you have an equation with a relatively simple u and dv, it might be better to use this method. Let's try it with this function.

Alright, let me explain this tricked-out diagram. Basically, one column shows u and its derivatives and the other shows dv and its anti-derivatives. Drawing those diagonal lines helps you keep track of which u multiplies with the dv, and putting the alternating signs above the lines tells you whether those will be positive or negative. So, x^2 will go with e^x and be positive, 2x will go with e^x and be negative, etc.

So, the final answer turns out like this.

Don't forget the + C.

If you want more practice, this cool little web page lets you solve a bunch of problems step by step in Flash.

http://archives.math.utk.edu/visual.calculus/4/int_by_parts.3/


REMINDER: Lisa, you're up next.

And now I finish my post with this Photoshop edit:


Hope this was helpful, folks. Until next time.

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