Mr. O'Brien's 2012-13 Period 4 AP Calculus

Friday, March 8, 2013

U-substitution practice

Practice problems with solution for u-substitution:

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/usubdirectory/USubstitution.html#PROBLEM 1

Wednesday, March 6, 2013

Mr. O'Brien is off at the math meet. Please do p. 390/25 and 27 in class from IW #2.

Since #25 makes use of the regression calculator, please refer to these videos from Period 2:

Part 1
Part 2

I suggest watching them as a class and then referring back to them if you get stuck doing a particular step.

Monday, March 4, 2013

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):
$\int_{a}^{b}f'(x)dx=f(b)-f(a)$
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 $f(x)=x^{2}$
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 $f(x)=x^{2}$ which we can figure out from the graph now to be $g(x)=\frac{1}{3}x^{3}-\frac{1}{3}$ . This is the first part of the FTC which says $G(x)=\int_{a}^{x}f(t)dt$
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
$\int_{1}^{x}t^{2}dt=\frac{x^{3}}{3}-\frac{1}{3}$ . 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:

$f(x)-f(-1)=\frac{x^{3}}{3}-\frac{-1}{3}=\frac{x^{3}}{3}+\frac{1}{3}$

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

Next we went into the difference between a definite integral: $\int_{a}^{b}f(x)dx$ vs. an indefinite integral: $\int f(x)dx$ . 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.) $\int \sqrt{cotx} csc^{2}xdx$

$\frac{du}{dx}=-csc^{2}x$

$du=-csc^{2}xdx$

*sub u and du into the original integral

$=-\int u^{1/2}du$

$=-\frac{2}{3}u^{3/2}+C$
*sub back in what u is equal to to get your answer

$=-\frac{2}{3}cot^{\frac{3}{2}}x+C$

34) $\int tan^{7}(\frac{x}{2})sec^{2}(\frac{x}{2})dx$

$let u=tan\frac{x}{2}$

$\frac{du}{dx}=\frac{1}{2}sec^{2}\frac{x}{2}dx$

$du=\frac{1}{2}sec^{2}\frac{x}{2}dx$
*sub in u and du into the original equation (make sure to multiply by 2 because of the 1/2 in the substitution

$=2\int u^{7}du$

$=2(\frac{u^{8}}{8})+C$
*sub back in what u is equal to to finish the problem
$=\frac{1}{4}tan\frac{8x}{2}+C$

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

$\frac{secx+tanx}{secx+tanx}$
*after multiply by the funny form of one we have:

$=\int \frac{sec^{2}x+secxtanx}{secx+tanx}dx$

$\frac{du}{dx}=secxtanx+sec^{2}x$

$du=(sec^{2}x+secxtanx)dx$

*sub in u
$=\int \frac{1}{u}du$

$=ln\left | u \right |+C$

*sub back in what u is equal to
$=ln\left | secx+tanx \right |+C$

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

Wednesday, February 6, 2013

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:

EXAMPLE:

Find the antiderivative of sinx: -cosx; and find the value for the antiderivative at the given interval:

Don't forget! Quiz next class on IW#1-8, IW#4-8 are also due on Friday! We will be having a text on Thursday (Valentines Day) so study up!

IW#9 is on:

Pg. 297/ Q. 1-4 (answers above)

Pg. 295/ Q. 19, 21, 25, 27, 33, 51

Pg. 206/ Q. 5, 9, 13, 23, 29, 45, 57, 61, 67

Good luck everyone!

P.S. I apologize for any spelling or grammatical errors, or just errors in general