1/15/11

Alrighttttt, this week we started to learn definite integrals. The definite integral of a function represents the area between the curve and the x-axis. *Definite integrals can also be calculated by counting squares and Riemann Sums. Sounds complicated? Not at all. Basically what you're doing is drawing the graph of the function, putting your points, drawing however many rectangles there asking for, looking to see which rectangle (right or left) touches the graph, or goes over, and plugging in..Or finding the area haha :)

Here's an example:

What are 2 types of Riemann sums that we could use to find the area of the region bounded by the graph of f(x)=1/2^2 and the x-axis between x=0 and x=3. Find the area ofthis region using all methods.
*So switching this to an integral we get the bounds of 3 as the upper, and 0 as the lower bound 1/2x^2=1/6x^3 bar 3 at top and 0 at the bottom.
*Now that we know 1/6x^3 is our formula were gonna plug in the bounds 3 and 0. 1/6(3)^3-1/6(0)^3=9/2 (this is usuing the fundamental theorm of calculus)

**Now draw the LRRAM
We know the function is a parabola, and the x axis goes up to 3, with 3 triangles, and the left point of the rectangles are touching the graph(which means it does NOT go over)
*Now plug in the 3 rectangle points into the original function (1/2x^2)
We're only going to plug in 0,1, and 2 because 3 is not touching the graph.
*1/2(0)^2=0
1/2(1)^2=1/2
1/2(2)^2=2
*Add them all together to get 5/2

**Now draw the RRAM
It's drawn the exact same except we know the rectangles go over, and when we plug in points were going to plug in 1,2, and 3.

That's pretty much all there is to it!

And now i'm off to state, goodluck sweetheartssssss! :D