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## AP Calculus BC 2008

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# AP Calculus BC exams: 2008 1 a

## Video transcript

I received a suggestion that I
do actual old AP exam problems, and I looked on the internet
and lo and behold, on the college board site, if you go
to collegeboard.com, you can actually get-- I couldn't find
the actual multiple choice questions, but you can find the
free response questions, and so this question is actually the
first free response question that they have on the calculus
BC that was administered just recently in 2008. So let's do this problem. And frankly, if you understand
how to do all of the free response questions, you
probably will do fairly well on the multiple choice, because
the free response tend to be a little bit more challenging,
especially the last parts of the free response. Well anyway, let's do this one. So I'll just read it out,
because I don't want to write it out all here, but this
is the actual diagram. I actually copied and pasted
this from the PDF that they provide on collegeboard.com. So it says, let r-- this is r--
be the region bounded by the graphs of y equals
sine pi of x. So let me write that down. So this top graph is y
is equal to sine pi x. and then the bottom graph is y
is equal to x cubed minus 4x. And how did I know that
this was the bottom one? Well I knew that this one
was sine of pi x, right? Because sine looks like this. It doesn't look
like that, right? When you go sine of pi
is 0, sine of 0 is 0, sine of 2pi is 0. So we do this as sine of pi x. Well anyway, they want-- so
this is the region between these two functions and part A
of this-- and this is kind of the softball question, just to
make sure that you know how to do definite integrals-- and
it says, find the area of r. So how do we do that? I think you know that we're
going to do a little definite integration, so let's do that. So then we're going to take the
definite integral, so let's just say the area is equal to--
I don't know if that's-- I hope I'm writing big enough for
you-- the area is going to be equal to the definite
integral from. So what are the x values? We're going to be going
from x is equal to 0 to x is equal to 2. And what's this? At any given point value of x,
what is kind of going to be the high-- when we're taking the
area, we're taking a bunch of rectangles that are
of dx width, right? So that's-- that's not dark
enough, I don't think that you can see that-- so that's
one of my rectangles. Whoops. Let's say that's one of my
rectangles right here that I'm going to be summing up. Its width is dx. What's its height? Its height is going to be
this top function minus this bottom function. So, essentially, we're going to
take the sum of all of these rectangles, so its height is
going to be-- let me switch colors arbitrarily-- the height
is going to be the top function minus the bottom function. So sine of pi x-- parentheses
here-- minus the bottom function. So minus x cubed plus 4x. Since I'm subtracting, I
switched both of these signs. And all of that times the width
of each of these little rectangles-- which is
infinitely small-- dx. And we're going to sum them
all up from x is equal to 0 to x is equal to 2. This should be fairly
straightforward for you. So how do we evaluate this? Well, we essentially take the
antiderivative of this and then evaluate that at 2
and then evaluate at 0. What's the antiderivative
of sine of pi x? Well, what functions
derivative is sine of x. Cosine of x-- let's see. If I were to take the
derivative of cosine-- let's say I took the derivative
of cosine pi x. This should be reasonably
familiar to you. Cosine of pi x, if I were
to take the derivative of it, what do I get? That equals pi. You take the derivative
of the inside, right? By the chain rule. So it's pi times the derivative
of the whole thing. The derivative of cosine of x
is minus sine of x, so the derivative to this is going to
be times minus sine of pi x, or you could say that equals
minus pi sine of pi x. So the derivative of cosine of
pi x is almost this, it just has that minus pi there, right? So let's see if we can rewrite
this so it looks just like the derivative of cosine pi x. And I'll switch to magenta. I want to make sure I
have enough space to do this entire problem. So let's write a minus 1
over pi times a minus pi. All I did, when you evaluate
this, this equals 1, so I can do this times sine pi x, and
then that's minus x to the third plus 4x, and then all
of that times the width dx. Well now we have it. We know that the antiderivative
of this is cosine pi x, right? And this is just
a constant term. So what's the antiderivative
of this whole thing? And I'll arbitrarily
switch colors again. The antiderivative
is cosine pi x. So we have minus 1 over pi
cosine pi x-- remember, I could just carry this over, this is
just a constant term-- this antiderivative is
this right here. And then these are a little
bit more straightforward. So minus the antiderivative of
x to the third is x to the fourth over 4 plus the
antiderivative of this is 4x squared over 2, or you could
just view that as 2x squared, and then we're going to
evaluate that at 2 and at 0, and let's do that. So this is equal to cosine of
2pi, and we'll have a minus sign out here, so minus cosine
of 2pi over pi, minus-- what's 2 to the fourth power? Let's see. 2 to the third is 8, 2 the
fourth is 16, 16 over 4 is 4, so it's minus 4, 2 squared is
4 times 2 is 8, so plus 8, so that's the antiderivative
evaluated at 2, and now let's subtract it evaluated at 0. So this will be minus cosine of
0 over pi-- all right, that's that evaluated at 0--
minus 0, plus 0. So these terms don't
contribute anything when you evaluate them at 0. And so what do we get? What's cosine of 2pi? Cosine of 2pi is the
same thing as cosine of 0, and it equals 1. What is the x value of the
unit circle at 2pi, or at 0? It's equal to 1. So this equals minus 1 over pi
minus 4 plus 8, and so this minus minus, those both become
pluses, cosine of 0 is also 1, so plus 1 over pi, and so this
minus 1 over pi and this plus 1 over pi will cancel out, and
all we're left with is minus 4 plus 8 and that is equal to 4. So that is part one, part A of
number one, on the 2008 DC free response questions. It actually took me a whole
video just to do that part. In the next video, I'll do part
B, and we'll just keep doing this, and I'll try to do a
couple of these every day. See you soon.

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