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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 11

Lesson 4: AP Calculus BC 2008# AP Calculus BC exams: 2008 1 c&d

parts c and d of problem 1 in the 2008 AP Calculus BC free response. Created by Sal Khan.

## Want to join the conversation?

- At5:42, I've put the exact same definite integral into my calculator (Casio fx-9860 GII) and it just doesn't seem to give me the same answer. I get approx. 10.227. I'm positive that I have the integral entered correctly and I'm positive that I understand the concept. Is there anything I can check to determine whether or not I have it right?(2 votes)
- I encountered the same problem while using my TI-84 plus. Turns out the calculator was in degree mode. That means that the input for the trigonometric functions (sin, cos, tan, etc.) would be in degrees. But when we do problems like these, we want the input to be in radians, the more fundamental unit. That way, the zeroes and critical points of the function 'sin(pi*x)' will align with rational values of 'x'. Just change your calculator to radian mode, and you will get the right answer. In the TI-84, you can do this by pressing (MODE) and selecting (RADIAN). Hope I helped!(2 votes)

- If you wanted to evaluate the ∫ [0,2] (sin(pi*x)-x^3+4x)^2 dx normally, it's not "impossible" or "stupidly long" to do, it does take some time, so in a test i'm not sure if you could do it, it depends i guess, anyway here goes nothing:

first expand out (sin(pi*x)-x^3+4x)^2

For the sake of knowing what's going on in there :

a=sin(pi*x), b=x^3, c= 4x

(a-b+c)^2 = (a-b+c)(a-b+c)= a^2 -2ab+2ac+b^2-2bc+c^2

if we subtitute back a b and c, we get:

V=∫ [0,2] sin^2(pi*x) -2x^3 sin(pi*x) +8x sin(pi*x) + x^6 -8x^4 +16x^2 dx

we'll split the integral to treat each parts separately :

A= ∫ [0,2] sin^2(pi*x) dx

-2B= -2 ∫ [0,2] x^3 sin(pi*x) dx (this one will be annoying)

8C=8 ∫ [0,2] x sin(pi*x) dx

D = ∫ [0,2] x^6-8x^4+16x^2 dx

So V = A-2B+8C+D

A= ∫ [0,2] sin^2(pi*x) dx

∫ [0,2] 1/2(1-cos(2pi*x)) dx (trig identities)

1/2 ∫ [0,2] 1-cos(2pi*x) dx

1/2 [x - sin(2pi*x)/2pi] [0,2]

1/2 [(2-0)-(0-0)] = 2/2 = 1

A=1

-----------------------------------------

B) Now the hard/long one, lots of integration by parts, the goal is to get "x^3" down to 1 using integration by parts:

B=∫ [0,2] x^3 sin(pi*x) dx

Integration by parts, u=x^3, v'=sin(pi*x)

=x^3 (-cos(pi*x)/pi) - ∫ 3x^2 (-cos(pi*x)/pi) dx [0.2]

Clean up, pull the constant out of the integral

= -x^3/pi cos(pi*x) + 3/pi ∫ x^2 cos(pi*x) dx [0.2]

Integration by parts, u=x^2, v'=cos(pi*x)

= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - ∫ 2x sin(pi*x)/pi dx ) [0.2]

Clean up

= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi ∫ x sin(pi*x) dx ) [0.2]

Integration by parts, u=x, v'=cos(pi*x)

= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi (x * -cos(pi*x)/pi - ∫ -cos(pi*x)/pi dx )) [0.2]

Clean up

= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi (-x/pi * cos(pi*x) +1/pi ∫ cos(pi*x) dx )) [0.2]

Evaluate ∫ cos(pi*x) dx

= - x^3/pi * cos(pi*x) + 3/pi ( x^2 sin(pi*x) / pi - 2/pi (-x/pi * cos(pi*x) +1/pi sin(pi*x)/pi)) [0.2]

Distribute

= -x^3/pi cos(pi*x) +3x^2/pi^2 sin(pi*x)+6x/pi^3 cos(pi*x) -6/pi^4 sin(pi*x) [0,2]

Making it this far without making any careless mistakes is a miracle, so check if you didn't miss anything :) Now for the fun part, it may seem daunting to evaluate but if we look carefully, we can see that when x = 0 wherever there's a x, the whole part is = to 0, and the only part without 'x's running around by themselves, we have sin(pi*x), which when x=0 is also equal to 0, so that's one less thing to worry about. Also wherever there's a sin(pi*x) if x=2, we have sin(2pi) and that's also equal to 0. Which means that we only have 2 parts to evaluate at 2, which is great !

= -2^3/pi * cos(2pi) + 6*2/pi^3 cos(2pi)

= -8/pi + 12/pi^3

Rewritten that gives us:

B= 12-8pi^2 / pi^3

I guess we deserve a drum roll here, we're not done but still !

-------------------------------------------------

C= ∫ [0,2] x * sin(pi*x) dx

Integration by parts, u = x, v' = sin(pi*x)

= x * (-cos(pi*x)/pi) - ∫ (-cos(pi*x)/pi) dx [0,2]

Clean & evaluate

= -x/pi cos(pi*x) + 1/pi ∫ cos(pi*x) dx [0,2]

= -x/pi cos(pi*x) + 1/pi (sin(pi*x)/pi) [0,2]

= -x/pi cos(pi*x) + 1/pi^2 sin(pi*x) [0,2]

When x = 0, everything is equal to 0, and when x = 2, the second part with sin(pi*x) is also equal to 0, so that leaves us with just :

-2/pi * cos(2pi)

C= -2/pi

------------------------------------------------------------

D = ∫ [0,2] x^6-8x^4+16x^2 dx

= (x^7)/7 - 8(x^5)/5 + 16(x^3)/3 [0,2]

= (2^7)/7 - 8(2^5)/5 + 16(2^3)/3

A cool thing to notice here is that we only have a bunch of power of 2s:

= (2^7)/7 - 2^3(2^5)/5 + 2^4(2^3)/3

= (2^7)/7 - (2^8)/5 + (2^7)/3

we can factor out 2^7

=(2^7) (1/7 - 2/5 + 1/3)

=(2^7)/(7*5*3) (3*5-2(3*7)+7*5)

=(2^7)/(105) (15-42+35)

=(2^7)/(105) (8)=(2^7)(2^3)/(105)=(2^10)/105

D=1024/105

Done!

--------------------------------------------------

All that's left now is to evaluate V,

Quick recap :

V= A-2B+8C+D

A=1,

B= 12-8pi^2 / pi^3

C=-2/pi

D=1024/105

V= 1-2(12-8pi^2 / pi^3)+8(-2/pi)+1024/105

= (105+1024)/105 + (-24+16pi^2)/pi^3 - 16/pi

= 1129/105 + (-24+16pi^2-16pi^2 )/pi^3

V=1129/105 - 24/pi^3

It doesn't take "forever" but in a test idk how long you've got, but that's just one part of the question. lol, good exercize nonetheless. There might be faster ways to do it, i don't know, and it may contain some errors, i checked but i'm only human! The final answer is correct tho.

Random bragging, V = 1 + 2^10/(3*5*7) - 4!/pi^3, it's cute!(2 votes) - how do you ad fraction with decimals(0 votes)
- I'm not sure I understand your question, but you would have to convert both of them to the same representation. That most likely means converting the fraction to a decimal, since the decimal is probably irrational. Then you can add them as normal numbers. That's more a question for the arithmetic playlist, though.(4 votes)

- bad web connection how do I fix this(0 votes)
- Check your router and your internet speed, or diagnose the problem, then follow the instructions.(2 votes)

## Video transcript

So we were doing part c of the
first problem on the Calculus BC exam, the free response
part, and so what, I'll re-read it, it says the region
r, this is the region r, is the base of a solid. I redrew the region r here, but
with a little perspective, so we can hopefully visualize
it in three dimensions. So it says for this solid, each
cross section perpendicular to the x-axis, right? There's two ways you can do a
cross-section, you could do them, you could cut this
way, but that would be parallel to the x-axis. We wanted to cut perpendicular
to the x-axis, or parallel to the y-axis, right? So if we cut it along
the line like that. So they say, each cross-section
perpendicular to the x-axis is a square. So I drew a couple
of squares here. That's one. So here, this would be the
base, and since we know the cross-section is the square,
the height has to be the same length as the base. Same thing here. So here, the height will be
really high, because this is kind of maybe our maximum point
in terms of the base width. But it's still pretty wide,
then it gets narrow again. So how do we figure out
the volume of this solid? Which is kind of
hard to visualize. And it's in some ways you
easier to visualize it than it is to draw, so you have
to give me some credit. But anyway, what we do is, we
take the area of each of these squares, and I drew
one of them here. You take you each of the area
of each of these squares, multiply them by a super small
change in x, and that we know, from everything we've learned
in calculus hopefully, that that super small change in x,
I'm trying to draw a little perspective, is dx. So if we multiply dx times the
area of this square, that is the volume of this kind of
part the entire solid. And if we were to sum up all of
these infinitely thin solids, we would get the volume
for the whole thing. So how do we do that? Well let's write our
integral expression. So what is the area of
each of these squares? Each of these cross-sections,
what is the area? Well the base is going to
be the difference between our two functions, right? This top function right
here, that was sin of pi x. And this bottom function, right
here, that is y is equal to x to the third minus 4x. So the base of the functions,
the base of this distance right here is going to be the
difference between the top function and the
bottom function. So each base is going to be sin
of pi x, minus this function, so minus x to the third, plus
4x, right switch the sign. Minus x to the third, plus 4x. So far this might look pretty
similar to part a, but what's the twist here? We want the area. The area of each of
the squares, not just this distance. So what's the area? It's going to be this distance
squared, so we have to square this entire thing. So that's the area of each
of these squares, and then we have to multiply them
times a little bit of dx. And that gives us the volume
for each of these, I guess, parts of the entire solid. And then what are our
boundaries of integration? Well it's the same
thing as in part a. This is 0, this is 2. So the boundaries are
pretty straightforward. So we essentially just have
to evaluate this now. And just like in part b, I've
first tried to evaluate this analytically, and you end up
getting a very, very, very hairy integral. Which you can do analytically,
you but you have to know some power reduction formulas in
trigonometry, you have to do some integration by parts, it
would probably use up all of the time on the AP Exam. So since they said that a
graphing calculator is required for some parts of the problem,
I say why not use our graphing calculator? Because the graphing calculator
is very good at numerically evaluating definite
integrals like this. So let's get out my
TI-85 emulator again. Here we go. I want you see the key strokes
so that you, ok turned on, exit out of this. So we're going to use the
calculus function here, so second, calculus. And this function right here,
this is definite integral, a very useful function to use. Press F5, definite integral. And then we just type
in the expression. So the expression, let me
move it down a little bit. So it's open parentheses
sin of where is pi? Pi is -- I haven't used one of
these calculators in a long, long time -- oh there it is. Second pi x sin of pi x minus x
to the third power, plus 4x. All of that squared. And then this definite integral
function, you have to tell it which is the independent
variable, or kind of, you know, what variable are we
integrating across, and that's the variable x. And then you just tell it the
boundaries of integration, and we're done. So integrate from 0, from x is
equal to 0, to x is equal to 2. If I haven't made a mistake,
I can hit enter, and let the calculator do the
rest of the work. Let's see what it ends up with. OK, 9. -- that's the answer,
that is the volume of this solid -- it's 9.9783 So you
could write that this is equal to 9.9783. And I'm pretty sure they want
you to use a calculator, because frankly, computing the
integral, that's kind of just chug math, you know, very
mechanical math, although it's pretty sophisticated, but
it would take you forever. But this is kind of, I think,
what they wanted you to do, set up the integral, recognize that
each of the squares, the area of each of the squares is just
going to be this distance, the distance between the
functions squared. And then you integrate
that from 0 to 2. Let's see how much
time I have left. I have a couple of minutes. So let's do part d. The region r models, let me,
paste, oh, I didn't want to do that, edit, undo, edit,
paste, there you go. OK, so I wanted to make
this a little smaller. So what does part d say? The region r models the
surface of a small pond, so that's the surface now. At all points r at a distance x
from the y-axis, the depth of the water is given by h
of x is equal to 3-x. So essentially 3-x is
the depth, right? So at this point of this pond,
the depth is just 3, right 3-0. And at this point, the
depth is 3-2 which is 1. So essentially the pond is
going to get shallower and shallower or as we go
further to the right. You could almost imagine
it, let me see if I can draw it again. So this is the sin function
with some perspective. This is the polynomial
function below it. This is, it's probably drawn
in, that's the x-axis. This is the y-axis. And so here, the depth of the
pond is given by the function h of x is equal to 3-x. So over here, the depth is 3,
so if I were to draw, so if I were to go straight down the
depth is, you know maybe it's 3. And the pond essentially gets
shallower and shallower as we go to the right. So how do we figure out
the volume of this? Over here, what is the depth? It's going to be 1, right, 3-2. This is x is equal to 2. So here the depth
is going to be 1. So if we take the cross-section
just along the x-axis, the depth is going to look
something like this, but then this is the top of it. I know that's kind of
hard to visualize. But anyway, how do we
figure out what the volume of this lake is? But actually, I realize I'm
pushing nine minutes, so I will continue this
in the next video. See you soon.