Mr. O'Brien's 2012-13 Period 4 AP Calculus: Scribe Post 2/6

The past couple days have been treacherous in AP Calc as Mr. O’Brien ditched us in our time of need. We were left to fend for ourselves in the land of integrals and average values and when he finally returned he did not waste a second to throw yet another new lesson at us and the news of a quiz on this Friday, February 8th, on IW#1-8 and a test next Thursday, Happy Valentines Day from O’B to all his suffering calc students! YAY!
As class today started terror struck as O’B discovered he DESPERATELY needs a hair cut *GASP* and we didn’t have a scribe for the day *AH*

grumpy O'B

but, when all was in despair O’B embraced his shaggy dog, yet still product filled hair, Hayley found his look alike, and I volunteered to be scribe. Oh joy, here we go . . .

happy O'B!

O'B's look alike, I'm pretty sure this is Joseph from one of those biblical movies...

Before I begin on what we actually did in class I’m going to post some things that will be helpful for the quiz on Friday if O’B let’s us use the scribe posts for notes. Here are some good rules and equations to know:

RULES FOR DEFINITE INTEGRALS:

1. Order of Integration:

2. Zero:

3. Constant multiple:

4. Sum and Difference:

5. Additivity:

6. Max-Min Inequality: if max-f and min-f are the maximum and minimum values of f on [a,b] then:

7. Domination:

The AVERAGE (MEAN) VALUE:

If f is integrable on [a,b], it’s average (mean) value on [a,b] is:

Onward to the beginning to class! We started class off with and ended up spending most on questions 1-4 on page 297 of our textbooks. These four problems are also part of IW#8 so if you don’t have the work and/or answers down already I suggest you do that!

1.

In question one we discussed how the f(x) in an integral is the height of the rectangles made in the Riemann Sum and the dx, or the change in x, is the base. This means that the functions multiplied by the change in x (base X height) equals a+2b. Now if we add 3 to the height, f(x)+3, we’re shifting the whole function up 3 and adding a rectangle to the area underneath the curve. This added rectangle has a height of 3 and a base of the change of x which in this case is b – a.

graph showing the added rectangle from changing f(x) to f(x)+3


math using Sum and Difference Definite Integral Rule to solve

FINAL ANSWER: (D) 5b-2a

In this question we had to take a similar approach as we did in number 1. When looking at the given expression and thinking about the previous worksheets and questions we’ve done with the Riemann Sum we can see that the “1/20” is the base of all the rectangles or the change in x and all the numbers in the parentheses are the changing heights. Identifying the base of the rectangles, or the change in x, should be one of the first things you do for Riemann Sum.

When thinking of what we just found out in an integral form that means the 1/20 becomes dx and all the different heights become f(x). There’s nothing out in front of the integral, everything in the expression has now been accounted for, so we can cross out all answer except (A) and (B). We know that the integral is going from 0 to 1 because the square rot of 20/20 = 1, so know we just have to decide if f(x) is the equation in (A) or (B). But, if we know the integral is only going from 0 to 1, which are the x-values, and if you plug in from 0 to 1 as x in the equation in (A) you only end up with √(1/20) not √(20/20). With the equation in (B) we can get √(20/20) so we know know that’s our answer because with this equation the change in x is 1/20 not 1.

graph of Riemann Sum of function

FINAL ANSWER: (B)

to solve use zero rule (see above)

FINAL ANSWER: (C) 2

Going off this equation we looked at the integral:

and O’B asked why this integral is negative. After a short discussion we realized it’s because the integral is going backwards from 2 to -2 so it’s a negative change in x.

Then we looked at:

and O’B asked why, if the last integral was negative, is this one positive. This is because it’s a negative height, for being below the x-axis, multiplied by a negative base, for going backwards from 2 to -2, which gives you a positive number.

a.)

b.)

from the equation given above we know that:

so when using the given interval and the equation we found in part a we get:

then we can use our calculators and plug in fnINT[f,x,a,b] and get an answer of -6. We then must multiply by1/2 so the final answer is -3.

BUT what if we can’t use our calculators O’Brien asks. Well, then we’d use The Fundamental Theorem of Calculus.

Let’s put the FUN in Calculus! The FUNdamental Theorem of Calculus! WOO!

But before we actually get into the two parts of the theorem let’s do a geogebra exercise to give us a visual of the Fundamental Theorem.

The end result should look like this:

The Fundamental Theorem of Calculus, Part 1:

If:

then:

Basically, the derivative of an integral is the function (f(x)). This is because taking the derivative of a function and taking the integral of a function are like opposites, they are inverses so they undo each other. This process is better explained my PatrickJMT in this video.

The Fundamental Theorem of Calculus, Part 2: