1/9/11

So this week we learned about definite integrals, I thinkkk this is somewheres in chapter four. Surprisingly it is very simple! Sooo to start off a definite integral is an integral with upper and lower bounds that represents an area between the graph and x-axis, bounded by numbers. So say there's an integral and the number at the top is b and the bottom is a. A would be the lower bound, or the smaller number. B would be the upper bound, or the larger number. **Note that it has to be this way to be correct, if it is the opposite and the larger number is at the bottom then you have to switch them by making the function negative, then switching the numbers.**Also, if both of the numbers are the same then the function will automatically be zero.

Example:
Sketch the region whose area is given by the intergral
Integral with 5 on top and 1 on the bottom x^2 dx
Now to start this off you would draw a graph, the x-axis would go to five, and the graph itself would be a parabola because in the function there is an x^2. After drawing the parabola you would shade from x=5 to till the x-axis reaches the parabola.

We also learned the fundamental theorem of calculus. Which is if theres an intergral with b on the top and a on the bottom, f(x)dx=f(b)=f(a)=area where f(x) is the integrated function.
It sounds complicating but here's an example.

Example:
Integral with 3 on the top and 1 on the bottom 4dx
Evaluate: First figure out what the derivative was to get 4, it would be 4x |^3 <-top number of integral. Then plug into the formula f(b)-f(a), so 4(3)-4(1)=8. Then graph, so make the x-axis up to 3, and the y-axis up to 4, then shade.

The only thing that I really find pretty confusing is all the AP stuff, I tried doing that binder thing and I don't really know what i'm doing. I guess it will be easier to understand once we go over it in class.