2/12/11

Soooooo...time for another blog I suppose. About what? Nothing specific really..All this week we just took practice AP's and did work with that. But anyway, I think I'll just give a few examples of some recent things we learned..

Ex. 1) S sin(3x+5) dx ...(integrate)
*Okay first you should notice that with this problem there's two things going on at once..kind of like a chain rule; so, that tells you that you're going to have to use substitution to solve this
*So your u is going to be what's inside the function, which is 3x+5 and your du will be 3 dx (because 3 is the derivative of your u)
*Now you're going to replace 3x+5 with u in your equation to solve...butttttt, you're going to have to add a 1/3 on the outside to even it out (since your derivative isn't actually in the original problem) So you should get this:
1/3 S sin(u) du
*Now you're going to integrate sin(u)....and you should get -cos(u)
So now you have this:
1/3[-cos(u)] + c
*Now all you have left to do is plug back in for u like this:
= 1/3[-cos(3x+5)] + c

Ex. 2) For y = 3x^2+2x+5 at the point (4,0), what is the slope of the normal line?
*First, since they're asking for slope you know that you're going to have to take the derivative of the equation they gave you.
*So once you take the derivative you should get 6x+2
*Since they didn't ask for an equation, all they want is the slope (just the number)..and they gave you an x-value so you're going to plug that (4) into your derivative (6x+2) like this:
6(4)+2
= 26...so that's your slope
*Buttttt, since they're asking for the slope of the NORMAL LINE, that means to get your final answer you'll have to take the negative reciprocal of the slope you just found
*So the slope of the normal line is -1/26

Ex. 3) Find the average value for f(x)=5x^2+6x+2 on the interval [0,2].
*Sooooo when you first look at this problem I bet the word "average" stands out to you..it probably makes you think like average rate of change or average velocity or something which means that you'll have to take the derivative....wellllll, not in this case. Average VALUE is totally different. It involves an integral, but it's super easy because all you have to do is use this simple formula:
1/b-a Sa,b f(x) ....(where its the integral of f(x) from bounds a-b multiplied by 1/b-a
*So plugging in the information they gave us into the formula you should get this:
1/2 S0,2 5x^2+6x+2 dx
*Now all you do is integrate it! And you should end up with this:
1/2[5/3x^3 + 3x^2 +2x] I2,0 ...(pretend that "I" is the bar that means you haven't plugged in yet, haha)
*So now do the Fundamental Theorem of Calculus:
1/2 [5/3(2)^3 + 3(2)^2 + 2(2)] - [0]
**^don't forget, the 1/2 is still out front because that's in the formula
*So once you simplify everything you should get this for your final answer:
44/3