0:00:01.030,0:00:05.980
I received a suggestion that I[br]do actual old AP exam problems,

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and I looked on the internet[br]and lo and behold, on the

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college board site, if you go[br]to collegeboard.com, you can

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actually get-- I couldn't find[br]the actual multiple choice

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questions, but you can find the[br]free response questions, and so

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this question is actually the[br]first free response question

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that they have on the calculus[br]BC that was administered

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just recently in 2008.

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So let's do this problem.

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And frankly, if you understand[br]how to do all of the free

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response questions, you[br]probably will do fairly well on

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the multiple choice, because[br]the free response tend to be a

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little bit more challenging,[br]especially the last parts

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of the free response.

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Well anyway, let's do this one.

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So I'll just read it out,[br]because I don't want to write

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it out all here, but this[br]is the actual diagram.

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I actually copied and pasted[br]this from the PDF that they

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provide on collegeboard.com.

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So it says, let r-- this is r--[br]be the region bounded by the

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graphs of y equals[br]sine pi of x.

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So let me write that down.

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So this top graph is y[br]is equal to sine pi x.

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and then the bottom graph is y[br]is equal to x cubed minus 4x.

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And how did I know that[br]this was the bottom one?

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Well I knew that this one[br]was sine of pi x, right?

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Because sine looks like this.

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It doesn't look[br]like that, right?

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When you go sine of pi[br]is 0, sine of 0 is

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0, sine of 2pi is 0.

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So we do this as sine of pi x.

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Well anyway, they want-- so[br]this is the region between

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these two functions and part A[br]of this-- and this is kind of

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the softball question, just to[br]make sure that you know how to

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do definite integrals-- and[br]it says, find the area of r.

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So how do we do that?

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I think you know that we're[br]going to do a little definite

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integration, so let's do that.

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So then we're going to take the[br]definite integral, so let's

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just say the area is equal to--[br]I don't know if that's-- I hope

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I'm writing big enough for[br]you-- the area is going to be

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equal to the definite[br]integral from.

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So what are the x values?

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We're going to be going[br]from x is equal to 0

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to x is equal to 2.

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And what's this?

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At any given point value of x,[br]what is kind of going to be the

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high-- when we're taking the[br]area, we're taking a bunch of

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rectangles that are[br]of dx width, right?

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So that's-- that's not dark[br]enough, I don't think that

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you can see that-- so that's[br]one of my rectangles.

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Whoops.

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Let's say that's one of my[br]rectangles right here that

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I'm going to be summing up.

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Its width is dx.

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What's its height?

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Its height is going to be[br]this top function minus

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this bottom function.

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So, essentially, we're going to[br]take the sum of all of these

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rectangles, so its height is[br]going to be-- let me switch

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colors arbitrarily-- the height[br]is going to be the top function

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minus the bottom function.

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So sine of pi x-- parentheses[br]here-- minus the

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bottom function.

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So minus x cubed plus 4x.

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Since I'm subtracting, I[br]switched both of these signs.

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And all of that times the width[br]of each of these little

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rectangles-- which is[br]infinitely small-- dx.

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And we're going to sum them[br]all up from x is equal

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to 0 to x is equal to 2.

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This should be fairly[br]straightforward for you.

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So how do we evaluate this?

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Well, we essentially take the[br]antiderivative of this and

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then evaluate that at 2[br]and then evaluate at 0.

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What's the antiderivative[br]of sine of pi x?

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Well, what functions[br]derivative is sine of x.

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Cosine of x-- let's see.

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If I were to take the[br]derivative of cosine-- let's

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say I took the derivative[br]of cosine pi x.

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This should be reasonably[br]familiar to you.

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Cosine of pi x, if I were[br]to take the derivative

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of it, what do I get?

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That equals pi.

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You take the derivative[br]of the inside, right?

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By the chain rule.

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So it's pi times the derivative[br]of the whole thing.

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The derivative of cosine of x[br]is minus sine of x, so the

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derivative to this is going to[br]be times minus sine of pi x, or

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you could say that equals[br]minus pi sine of pi x.

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So the derivative of cosine of[br]pi x is almost this, it just

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has that minus pi there, right?

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So let's see if we can rewrite[br]this so it looks just like the

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derivative of cosine pi x.

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And I'll switch to magenta.

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I want to make sure I[br]have enough space to do

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this entire problem.

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So let's write a minus 1[br]over pi times a minus pi.

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All I did, when you evaluate[br]this, this equals 1, so I can

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do this times sine pi x, and[br]then that's minus x to the

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third plus 4x, and then all[br]of that times the width dx.

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Well now we have it.

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We know that the antiderivative[br]of this is cosine pi x, right?

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And this is just[br]a constant term.

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So what's the antiderivative[br]of this whole thing?

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And I'll arbitrarily[br]switch colors again.

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The antiderivative[br]is cosine pi x.

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So we have minus 1 over pi[br]cosine pi x-- remember, I could

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just carry this over, this is[br]just a constant term-- this

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antiderivative is[br]this right here.

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And then these are a little[br]bit more straightforward.

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So minus the antiderivative of[br]x to the third is x to the

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fourth over 4 plus the[br]antiderivative of this is 4x

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squared over 2, or you could[br]just view that as 2x squared,

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and then we're going to[br]evaluate that at 2 and at

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0, and let's do that.

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So this is equal to cosine of[br]2pi, and we'll have a minus

0:07:03.510,0:07:09.930
sign out here, so minus cosine[br]of 2pi over pi, minus-- what's

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2 to the fourth power?

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Let's see.

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2 to the third is 8, 2 the[br]fourth is 16, 16 over 4 is 4,

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so it's minus 4, 2 squared is[br]4 times 2 is 8, so plus 8, so

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that's the antiderivative[br]evaluated at 2, and now let's

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subtract it evaluated at 0.

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So this will be minus cosine of[br]0 over pi-- all right, that's

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that evaluated at 0--[br]minus 0, plus 0.

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So these terms don't[br]contribute anything when

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you evaluate them at 0.

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And so what do we get?

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What's cosine of 2pi?

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Cosine of 2pi is the[br]same thing as cosine

0:08:01.110,0:08:03.090
of 0, and it equals 1.

0:08:03.090,0:08:06.490
What is the x value of the[br]unit circle at 2pi, or at 0?

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It's equal to 1.

0:08:07.070,0:08:15.670
So this equals minus 1 over pi[br]minus 4 plus 8, and so this

0:08:15.670,0:08:19.900
minus minus, those both become[br]pluses, cosine of 0 is also 1,

0:08:19.900,0:08:25.840
so plus 1 over pi, and so this[br]minus 1 over pi and this plus 1

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over pi will cancel out, and[br]all we're left with is minus 4

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plus 8 and that is equal to 4.

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So that is part one, part A of[br]number one, on the 2008 DC

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free response questions.

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It actually took me a whole[br]video just to do that part.

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In the next video, I'll do part[br]B, and we'll just keep doing

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this, and I'll try to do a[br]couple of these every day.

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See you soon.