1/21/11

Alright so this week we continued with the UNO study and I think the material is actually really simple now. Last week we learned how to find LRAM, RRAM, how to use the Trapezoidal Rule, and how to find the exact value of the area of a region using the Fundamental Theorem of Calculus. This week we learned MRAM, which is the Midpoint Rule. This rule uses rectangles that touch the graph at the midpoint of the rectangle instead of the left or right side. The only thing different that you have to do for this is plug the midpoint of each rectangle into the original equation to get your heights. And to find the midpoint of each, all you do is add the numbers on each side of the rectangle and divide by 2. Delta x would simply be the distance between each rectangle. And after you get the heights, all you do to find the area is multiply each height by delta x and add them all together....This week we also learned how to find the area between 2 curves. This is when you're given two separate equations: you first graph them on the same coordinate plane; then you have to decide which curve is on top; then you use this formula: Stop-bottom (where you subtract the two equations and then integrate it). After that (if you aren't given the bounds in the directions), to find them all you do is set the two equations equal to each other and solve for x. Then once you integrate, you plug in the numbers of the bounds into the area formula ("Fundamental Theorem of Calculus"). And that's it! So easy!
So here's an example:

1.) Find the area between the two curves for y = x^2 + 2x + 1 and y = 2x + 5.
*Okay so the first thing you have to do is draw this graph so that you can figure out which one of these curves is on top.
*So the graph of x^2+2x+1 is a parabola with vertex at x=-1 opening up...And the graph of 2x+5 is a line going up intersecting the parabola at points (-5/2,0) and (0,5). So that means that the line is on top and the parabola is on the bottom..And the area in between those two graphs is what you're going to be finding.
*But first, let's find our bounds since we were not given them...So to do that you just set your two equations equal to each other and solve for x like this:
x^2+2x+1 = 2x+5
x^2-4 ...(subtracted 2x and 5 over to the left)
(x+2)(x-2) ...(difference of two squares)
x=2, -2 And these are your bounds! 2 going on the top, and -2 on the bottom
*Now we can use the formula to find the area. We already said that the line 2x+5 is on top so here's how to set up the equation:
S(2x+5)-(x^2+2x+1)
Now distribute the negative and simplify it to get this:
S -x^2+4
Now integrate that and you should end up with this: (before plugging in your bounds)
-1/3x^3+4x
Now you can use the Fundamental Theorem of Calculus to plug in your bounds and find the area. So this is what you should get when you plug in:
-1/3(2)^3+4(2)-[-1/3(-2)^3+4(-2)]
= 32/3 ..And that's your area!

**Soooooooooo, this week was really good; I understood everything we went over and honestly this material is extremely easy!