Weeks 14 and 16

Week 14:

In week 14, we learned many things. We learned substitution, area between curves, e integration, and integration with natural logs.

Substitution needs to be done when there is a product, quotient, or chain rule in an itegral. We need to use substitution because product, quotient, and chain rules cannot be used in integration. Substitution is basically forcing a problem by any means to work. In substitution problems, there will be a derivative and a non-derivative usually multiplied or divided by each other. U will be set equal to the non-derivative and du will be set equal to the derivative. In most of the simple probles, the derivative is exact in the problem, but if it isn't, you would need to put the reciprocal of whatever is needed on the inside on the outside of the integral. The steps for substitution are as follows:

1. Find a derivative inside the integral

2. Set u= to the non-derivative

3. Take the derivative of u

4. Substitute u back in

If there is an extra variable in the derivative, it cannot be taken out by using the reciprocal on the outside. You would have to set u=0 and solve for the variable. You would then put this in the place of the variable and solve the same way you would solve any other substitution problem thereafter.

There is also substitution with e and natural log. Many people see e in their problems and automatically think the problem will be hard, but e integration is the exact opposite. In e integration, u will always be e's exponent. Du will be the derivative of u, which will most likely be the other term in the problem. If it is not, you would fix this just as in regular substitution either with putting the reciprocal outisde of the integral, or solving u for the variable.

Natural log integration is also very simple. Natural logs are found in fractional integrals. U will be the bottom, with du being the top. Once this is found, you plug u into: ln u + c



Week 16:

Last week, we learned trig inverse integration, how to find volume by using disks, and by using washers. The term disk is used when there is no hole in the volume, and the term washer is used when referring to a volume with a hole.

For trig inverse integration there are four different formulas:

sr = square root

1. S du/sr(a^2-u^2) = 1/du(arcsin(u/a)+c)

2. S du/a^2+u^2 = 1/dua(arctan(u/a)+c)

3. S du/u(su(u^2-a^2)) = 1/dua(arcsec(u/a)+c)

Most people get confused with this, as I do. The biggest problem people make is they don't square u and a after they put them into the equation.

Volume by disks is used to find the volume of a solid object

The formula for this is: pi S a-b [R(x)]^2 dx

For example, if you are given a problem that asks you to find the volume of a solid by rotating the region bounded by f(x)=2-x^2 on the x axis from 0 to 2

You would first draw the graph. This graph would be a curve and it would be rotated about the x axis, making it a solid, circular-shaped figure. You then just plug it into the formula and solve.

For volume by washers:

I personally think washers is easier. Maybe it is because I know how to do area under a curve, and this is what I relate it to. The formula for this is:

pi S a-b top^2-bottom^2 dx
*you will always have two equations.

For washers, you also graph the equations, but this time to see which curve is on top. The curve on top will be the first equation in your problem. The one on the bottom will be the one you'll be subtracting by. This is why I relate it to area under a curve.

For the things I do not understand:
This week was pretty fast paste, and it seemed like once we went over one thing, we were moving straight on to the next, and I was left completley lost. I understand the formulas and how to plug into them for trig inverse and disks, but I do not know where to go from there. Can someone please help me find some sort of steps to go by for this?