We started with:

This was easily found with the use of a calculator. But if not given use to a calculator, it's good to know how to solve this question.

We found the antiderivative of the given function over the interval of 0 to pi and replaced all the values of x with the values of the interval to find the difference in area. Now if no calculator is given, it's simple, just simply simplify it to the simplest form. Then repeat for the other question.

Our next question had to deal with the same functions as the last, but we have to find the integral equation of the volume of the shape over the same interval of 0 to pi, but when it's generated over the x-axis, and then the we found the volume when generated over the y-axis.

Now we cut this shape into tiny cylinders and find the volume of each cylinder, given by V = (pi)r

^{2}h but because we are dealing with 2 different functions, and finding the difference in the volumes, each graph has its own radius. The volume is then simply stated as V = pi(R

^{2}- r

^{2})h, we use an infinite amount of cross sections, and because of this the height is quiet small, its so small that we call it dx. Now we throw the functions' radius' into the equation and voila.

When revolving around the y-axis, it's the same deal, but with the use of a different equation. Note: I'm not sure what it is but I'll edit it in when I find out.

Our next question involved trapezoid sums.

The question shows a table with 9 sets of information when x is various values. We need to find the difference of area between the 2 given functions. Using a trapezoidal approximation, we find the upper limits and lower limits of f(x) and find the differences between these limits of g(x) and find the average. Quite simple.

Next question deals with finding the area of the region of a graph.

Now because the graph has a root between the given interval, we split the region into 2 seperate intervals. One finding the area of the region between the root at -1.100457 and -1.5 (I'm not sure why we're finding the area seemingly backwards on the interval) and we add this to the area of the function in the interval -1.100457 to 1.5

This is all we did in today's class. Review questions are in today's slides. Our exam is in 2 days and good luck to everyone. Next scribe will be .:. J + ME .:. or better known as Jamie.

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