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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 2: AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

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# AP Calculus BC exams: 2008 1 d

Part 1d of the 2008 AP Calculus BC exam (free response). Created by Sal Khan.

## Want to join the conversation?

- how would you integrate that?(3 votes)
- It cannot be broken up into separate functions when it is written like (sin(pi*x) - (x^3 - 4x))*(3 - x) because this is technically just one big term. (Terms are things separated by a "+" or a "-".) You can simplify and multiply it out though:

(sin(pi*x) - (x^3 - 4x))*(3 - x)

= (sin(pi*x) - x^3 + 4x)*(3 - x) [now distribute...]

= sin(pi*x)(3 - x) - (x^3)(3 - x) + 4x(3 - x) [you could have distributed another way, but you should get the same answer]

= 3sin(pi*x) - xsin(pi*x) - 3x^3 + x^4 + 12x - 4x^2

Now we have 6 terms, which can all be done separately, but I would probably group them like this:

[-xsin(pi*x)] + [3sin(pi*x) + x^4 - 3x^3 - 4x^2 + 12x]

Then the group on the right is easy to integrate, and the group on the left [-xsin(pi*x)] is pretty straightforward to integrate by parts.(11 votes)

- Does the AP exam question specify to how many digits to the right of the decimal point we should round our answer?(3 votes)
- Wouldn't you have to put units for the volume?(1 vote)
- only if they specify units on the AP exam, if they don't you can just right "x units" or not including anything besides the answer(4 votes)

- Were the equations of the two functions provided?(1 vote)
- Yes sure, they were shown in the previous videos(1 vote)

## Video transcript

Welcome back. We're doing the last part
of Problem 1 of the 2008 Calculus BC exam. And I'll repeat the problem. It says, the region R models
the surface of a small pond. At all points in R at a
distance x from the y-axis, the depth of the water is given by
h of x is equal to 3 minus x. Find the volume of the
water in the pond. So we thought about, this is,
you know, a perspective drawing of this, this is the surface
of the pond, right here. And I was trying to
draw the depth. So when x is equal to 0, the
pond is exactly 3, I guess, you know, we could say feet, or
whatever the units are, deep, and then when x is equal to
2, the pond is 1 unit deep. And so to find the volume, we
think about it the same way. What does a cross section
at any point of the pond look like? Well if we, let's see. So if we were to take a cross
section, so let's say that this, let's take this cross
section of the pond. Let me do it in a different
color, I know it can be kind of confusing. That's not too different. So that's green. So the height along the surface
of the pond is going to be the difference between
the two functions. And we know that the top
function is sine of pi x, and we know that this bottom
function is x to the third minus 4x. So if I were to draw, let
me see if I can draw this cross section. So we know this the height,
sorry, the width along the surface of this cross section
is going to be the difference between this function and
this function, right? So if I were to draw, let me
see if I can, so this is that same cross section, it's the
width along the top, I know I'm confusing you, is going to be
this function minus this function. It's going to be sine of pi x
minus x to the third minus 4x. And what's going to be
the height, right here? Well, they told us. The height of the pond, or
the depth of the pond, at any point, is 3 minus x,
whatever x value we're at. So this is going
to be 3 minus x. So the area of this cross
section of the pond is going to be sine of pi x minus x to the
third minus four x times the depth of the pond, right? That's the surface
width times the depth. So that times 3 minus x. So that's the area of
each cross section. So we want the volume for
the entire pond, we take essentially the volume of
each of these kind of slivers of the pond. So we take the area of each
cross section, and we multiply it by a very small width to
get a very small sliver. So we take that and multiply
it by dx, and we integrate. We sum up all of these slivers
of the pond from x is equal to 0 to x is equal to 2. So let's do that. Let's integrate from x is equal
to 0 from x is equal to 2. Once again, this is a
really hard integral. Or it's something that you can
do, especially if you use integration by parts, but it's
a little bit messy, and you only have 45 minutes to do all
three problems on the BC exam, so I'm assuming that they want
you to use the calculator here. So let's use the calculator
to evaluate it. So let's see. So this was part c that we
already have in our calculator. We could do second enter, and
we'll get the previous entry. Second enter. Because we already typed
in a bunch of stuff. The only difference between
what we did in Part C and now is, now, instead of having in
Part C, we had this expression squared. Now, we don't have it squared,
but we're multiplying it by times 3 minus x. So let's do that. So if we go here, we can
delete this squared. We're not squaring it anymore. Deleted. And now let's do, I want
you to see my key strokes. So let's do second insert. So let's insert times 3
minus x close parentheses. And so, let's see. We have sine of pi x minus x
to the third plus 4x, right, that's the same thing as this. Times 3 minus x. And our variable of integration
is x, and we're integrating from x is equal to zero
to x is equal to 2. Let's hit enter. And it noticed, it
is calculating. And our answer is 8.37, I
think, is a fair answer. So the volume this time,
the volume of the pond, is equal to 8.37. Anyway, hopefully you
found that useful. I will, every day, try to do
a couple of these problems. See you in the next video.