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### Course: AP®︎/College Calculus AB > Unit 9

Lesson 3: AP Calculus AB 2011 free response- 2011 Calculus AB free response #1a
- 2011 Calculus AB Free Response #1 (b, c, & d)
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

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# 2011 Calculus AB free response #3 (a & b)

Equation of a tangent line and area between curves. Created by Sal Khan.

## Want to join the conversation?

- Don't you have to write units squared right after the -1/8 + 1/pi because you are finding the area?(4 votes)
- Unless units are given, you don't have to add a generic 'units^2' to your answer on the AP exam. That said it doesn't count against you if you do add it.(4 votes)

- 4:55Where do these trig function properties of derivatives/antiderivatives come from? I'm having trouble grasping how to take the derivative of functions in which x is being multiplied by a constant, such as sin(2x), cos(3x/4) etc.

I perused the obvious vids on here, but didn't see any mention of these types of derivatives/antiderivatives.

thanks in advance(5 votes)- I found this on wolfram alpha but I'm still not sure what is going on in the u substitution step.

Possible intermediate steps:

integral sin(pi x) dx

For the integrand sin(pi x), substitute u = pi x and du = pi dx:

= 1/pi integral sin(u) du

The integral of sin(u) is -cos(u):

= -(cos(u))/pi+constant

Substitute back for u = pi x:

= -(cos(pi x))/pi+constant

Is this saying to multiply the entire integral by pi/pi? So just using a little trick of multiplying it by 1 and hence not changing anything?

(1 vote)

- at3:10how do you know which function is above/below?(3 votes)
- Usually the best way is to make a graph of the region and determine the answer by looking at the graph. In this case they've given you the graph, so you just have to recognize which line looks like part of the sine function (sin(πx) looks like sin(x) except for greater frequency) and which line looks like x^3 (again, there's a coefficient, but that doesn't change the general shape of the function).

If you're unsure, you can figure the slope of each function at the left boundary (x = 0), and the one with the higher slope would be the one that's above the other. In a pinch you can try to find a point within the range where you can calculate the value of both functions. By far the most valuable skill here, though, is to understand the graphs.(4 votes)

- @3:46Why are we evaluating/integrating from 0 to 1/2 instead of 0 to 1?(2 votes)
- Because we are integrating with respect to x, and the x values in consideration range from 0 to 1/2. We could change the integration to be with respect to y, in which case you would integrate from 0 to 1.(4 votes)

- I see that the derivative of -1/pi(cos-pi-x) is sin-pi-x, but going the other way is not overtly obvious. What is the operation for taking the integral of sin-pi-x(2 votes)
- If you know how to do u-subs, then make u=pi*x, and du=pi*dx. If you know what to do from here, great. Otherwise, let me know.(2 votes)

- is there any quick and easy ap cal ab crash course?(1 vote)
- Not on Khan Academy, but you could use parts of this course to help you study/learn.(2 votes)

- Where did Sal get that he couldn't use the calculator to evaluate the integral?(2 votes)
- I thought it was 1-4 are no calculator, and 5 and 6 are calculator-active.(1 vote)

## Video transcript

Let R be the region in the
first quadrant enclosed by the graphs of f
of x is equal to 8x to the third and g of x
is equal to sine of pi x, as shown in the figure above. And they drew the
figure right over here. Part a, write the equation
for the line tangent to the graph of f at
x is equal to 1/2. So let me just
redraw it here just so that-- I like drawing on
the black background, I guess, is the main reason
why I'm redrawing it. So the function f of x is
equal to 8x to the third, looks like this. It looks like that. This is our f of x-axis,
and this is our x-axis. And we want the equation
for the line tangent at x is equal to 1/2. So this is x is equal to 1/2. If you go up here, if
you evaluate f of 1/2, you get 8 times
1/2 to the third, which is 8 times
1/8, which is 1. And they actually gave us
that already on this point. This is the point 1/2 comma 1. And we need to find the
equation for the tangent line. So the tangent line will
look something like that. And to figure out
this equation, we just need to figure out
its slope, and then we know a point that it's on--
and you could use point-slope or you could just use your kind
of standard slope-intercept form to give an
equation for that line. So the first part, let's
figure out its slope. And the slope of
the tangent line is going to be the
same slope as the slope of our function at that point. Or another way to
think about it, it is going to be
f prime of 1/2, or the derivative
evaluated at 1/2. The derivative
gives us the slope of that line at any point. So what is f prime
of x? f prime of x is just the derivative of this. So 3 times 8 is 24
times x squared. 24 x squared. f prime of 1/2 is equal
to 24 times 1/2 squared, which is equal to 24 times
1/4, which is equal to 6. So the slope of this
line is equal to 6. I'll use m for slope. That's the convention
that we used when we first learned
it in algebra. So the slope is going to be 6. So the general
equation for this line is y is equal to mx plus b. This is the slope. This is the y-intercept. We already know
that the slope is 6, and then we can use the
fact that the line goes through the point 1/2, 1
to figure out what b is. So when y is 1, 1 is equal to
our slope times x. x is 1/2. Or another way to say
it, when x is 1/2, y is 1 plus some y-intercept. Or if I take x is
1/2, I multiply it times the slope plus the
y-intercept, I should get 1. And so I get 1 is
equal to 3 plus b. I can subtract 3 from both
sides, and I get negative 2 is equal to b. So the equation of the
line is going to be y is equal to 6x minus 2. That is the equation
of the tangent line. Now part b, find
the area of R. So R is this region right over here. It's bounded above
by g of x, which they've defined as sine of pi x. It's bounded below by f
of x, or 8x to the third. So the area is going
to be-- actually, let me just do it-- I'll
scroll down a little bit. I still want to be able to see
this graph right over here. Part b. The area of R is going to be
equal to the integral from 0, that's this point of
intersection right over here, to 1/2. So let me make it clear. This is 0 to 1/2. And then the
function on the top-- so we could just take
the area of that, but then we're going to
have to subtract from that the area underneath
the function below. Or one way to think about
it is the integral from 0 to 1/2 of the top function is
g of x, which is sine of pi x. But if we just evaluated
this integral-- let me just put a dx over here. If we just evaluated
this, we would get the area of
this entire region. But what we need to
do is subtract out the area underneath the
second, underneath f of x. We just subtract out
the area under that. So we just subtract
from that f of x. And f of x, we already
saw, is 8x to the third. Is 8x to the third power. And now we can
just evaluate this. So let me draw a
little line here. It's getting a little bit messy. I'll just do it down here. So we need to take the
antiderivative of sine of pi x. Well, the derivative of cosine
of x is negative sine x. The derivative of cosine of pi
x is negative pi cosine of pi x. So the antiderivative
of sine of pi x is going to be negative
1 over pi cosine of pi x. And you can verify
it for yourself. And you might say, wait. Sal, how did you know
it was a negative. Well, I put the
negative there so that when I take the derivative
of the cosine of pi x, I would get a negative sign, but
that negative will cancel out the negative to give
me a positive here. And you say, why do you
put a 1 over pi here. Well, when you take the
derivative of this thing using the chain rule, you take
the derivative of the pi x, you'll get pi, that you
would multiply everything by, and then you would get
negative sine of pi x. And that pi doesn't
show up here. So I need something for
it to cancel out with. And that's what this 1 over pi
is going to cancel out with. And you could do use
substitution and all the rest, if you found something
like that useful. But it's in general,
a good habit or I guess it's good to be able
to do this almost by sight. So, and you can verify
that this derivative is equal to sine of pi x. So the antiderivative
of sine of pi x is this. The antiderivative of
negative 8x to the third power is negative 8-- I'm going to
divide it by 4-- so negative 2x to the fourth power. And all I did is I
incremented the 3 to a 4, and then I divided
by 8 by the 4. And you could take
the derivative of this to verify this it is the
same thing as negative 8x to the third power. And we're going to have to
evaluate that from 0 to 1/2. When you evaluated 1/2,
I'm going to get negative 1 over pi cosine of pi
over 2 minus 2 times 1/2 to the fourth power is 1/16. So that's evaluated at 1/2. And then from that, I'm
going to subtract negative 1 over pi cosine of 0. Let me just write it out. So minus negative 1 over
pi cosine of pi times 0. Let me just write
cosine of 0 pi, I could write, or pi times 0,
minus 2 times 0 to the fourth. So that's just
going to be minus 0. So let's evaluate this. So to simplify it, we have
a cosine of pi over 2. Cosine of pi over 2
is just going to be 0, so this whole thing
just becomes zero. And then you have a
negative 2 divided by 16. That's negative 1/8. And then from that, I'm going
to subtract this business here. Cosine of 0, this is 1. So this is just a
negative 1 over pi. And then I have a 0 there. so I can ignore that. So this is equal to negative
1 over 8 plus 1 over pi. And we are done. This part of it, you're not
allowed to use a calculator. So this is about
as far as I would expect them to
expect you to get. And so I'll leave you there. In the next video,
we will do part c.