Scribe Post 3/1

We started off class reviewing all the material we learned throughout the year and what we have left to learn. Something important Mr. O'Brien said to note was this statement on the Fundamental Theorem of Calculus: "The definite interval of the rate of change of a quantity over an interval interpreted as the change of the quantity over the interval."(FTC):

Our new material today was "Anti-derivative by substitution of variables (including change of limits for definite integrals)"
We started our lesson by going on geogebra and making an integral (instructions below):
1. plot
2. let a=1 and make a slider
3. put a point A on the x-axis
4. type into the input: Integral[f,a,x(A)]
This is what it should look like so far:

5. plot the point (x(A),b)
(This should make a point B show up on the graph)
6. turn trace on for point B
7. move A up and down the x-axis
This is what you should see now:

The function shown in the black traced by point B is the anti-derivative of which we can figure out from the graph now to be . This is the first part of the FTC which says
G(x) being an anti-derivative of f which you can find using a constant to a variable which is what we did in geobebra above
  . If you change the constant from 3 instead of 1 (change the slider on geogebra from 1 to 3) the C value changes, in this case from -1/3 to -9. We were asked what we thought the C value would be if we changed the constant to -1 and someone guessed 1/3 which was correct! We checked our answer using the FTC:



*Note: On page 337 there a bunch of anti-derivative formulas if needing help 

Next we went into the difference between a definite integral: vs. an indefinite integral: . A definite integral is an accumulation like an area while an indefinite integral is an anti-derivative.

Mr. O'Brien did an example on the board which was to show us that the variable of integration (dx, du, whatever it may be) is important. And in order to be able to solve an integral all the variables have to be the same. We needed to know this going into integration by substitution because when substituting you have to make sure that all the variables are the same before you can solve. See example 3 on page 338.

Finally we started working on INTEGRATION BY SUBSTITUTION by doing examples from IW #1 on page 342

32.)



 

 

*sub u and du into the original integral



 
*sub back in what u is equal to to get your answer



34)  




 


 *sub in u and du into the original equation (make sure to multiply by 2 because of the 1/2 in the substitution



 
*sub back in what u is equal to to finish the problem


45)
*the trick with this problem is to multiply by a funny form of one:

 
*after multiply by the funny form of one we have:

 

 

 

 

*sub in u
 

 

*sub back in what u is equal to  
 

Problems 48 and 54 from IW #1 will also be finished in class, we just ran out of time to finish them today.
Just in case anyone wants to start revising for the AP Calculus test early... Mr. O'Brien suggest using  Kahn Academy

IW #1
p. 342/32, 34, 45, 48, 54 (done in class)
p. 342/5, 17, 21, 25, 27, 33, 39, 47, 55, 59, 71, 73, 76, 79