AP Calculus BC exams: 2008 1 a Part 1a of the 2008 BC free response
AP Calculus BC exams: 2008 1 a
- 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.
- 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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