Post #14

This week in calculus, we learned average value, substitution, e integration, and ln integration. Finding an average value is the same steps as finding a definite interval except you have to multiply the integral by 1/b-a. Example:

Find the average value of f(x)=x on [0,3]

1/ 3-0 = 1/3

1/3[1/2x^2]

then plug in 1/3 [ 1/2 (3)^2 - 1/2 (0) ^2] = 3/2

ln integration is really simple, you just have to recognize it. For ln integration, the top has to be the derivative of the bottom of the fraction.

Examples:

the integral of 1/x dx= ln |x| + c

the integral of 4/x dx = 4 ln |x| +c

the integral of 1/4x-1

u = 4x-1 du= 4 dx

1/4 the integral of 1/x du = 1/4 ln |4x-1| + c

Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:

1. Find a derivative inside the interval

2. set u = the non-derivative

3. take the derivative of u

4. substitute back in

Example:

(x^3 +1) (3x^2) dx

u= x^3 +1 du= 3x^2 dx

integral of u du

1/2 u ^2 + c

1/2 (x^3 + 1) ^2 + c

I think I am mostly confused with substitution with variables. I usually get lost after I solve for x. I don't know what to do next or what to plug in to. I am still having trouble with graphs so I am struggling sketching the graph for area between curves, but if the graph is sketch I know how to solve it so if anyone can help me with graphing without using my calculator please do.

Hope you all have a good holiday!