2/6/11

Sooo..starting off this week we learned the Second Fundamental Theorem of Calculus. All that is is the derivative of an integral. So it would be written in the form: d/dx S1,2x x^2 dx...where d/dx infers derivative; 1,2x are the bounds; and x^2 is the equation. In order to solve a problem like this, all you do is take f(x) (which would be the equation), plug in the top bound for x and multiply that by the derivative of the top bound. *Also, in order for this method to work, the bottom bound has to be a number. Nexxttttt, we learned integration by Substitution..you use this method when you have an equation that looks like it would be a product rule or a quotient rule. To use substitution, one part of the equation is going to be the derivative of the other part..or close to it. It's a weird process to explain so I'll just give an example haha...And also this week we learned how to integrate e..and all you do for that is basically the opposite that you would do to take its derivative. First you recopy the e part, then instead of multiplying by the exponent's derivative, you divide by it. So here are a few examples of what I just talked about:

Ex. 1) d/dx S6,2x^4 2x+1 dx......where 6,2x^4 are the bounds and 2x+1 is the equation
*So first you're going to take the equation (f) and plug in the top bound for x..so you should get this:
2(2x^4)+1
*Now multiply that by the derivative of 2x^4...which is 8x^3
*So you should have 8x^3(4x^4+1)

Ex. 2) S(6x)(3x^2+1) dx
*Now since this looks like it would be a product rule, you have to check to see if you can use substitution to solve it
*6x is the derivative of 3x^2+1..so yes you can use substitution
*To use this method, you're going to replace one of the terms with u to make it simpler to solve
*Since 6x is the derivative of 3x^2+1, u becomes 3x^2+1 And du, the derivative of u, would be 6x dx..(<*So now once you replace 3x^2+1 with u, you're simpler integral should look like this:
S u du (and the 6x is not included because it's going to disappear)
*So once you integrate u, you should end up with this:
1/2u^2 + c ...*don't forget the +c because there are no bounds
*Now all you do is plug back in for u. So you're final equation should look like this:
1/2(3x^2+1) + c

Ex. 3) S e^4x^2
*To integrate this, you first recopy the equation:
= e^4x^2
*Now all you do is divide by the derivative of e's exponent...which is 8x
*So you should get this for your answer:
e^4x^2/8x

**Well this week was good overall. I understood everything; I just think I could use more practice on the substitution problems because those are kinda tricky.