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## AP®︎ Calculus BC (2017 edition)

### Course: AP®︎ Calculus BC (2017 edition) > Unit 9

Lesson 12: Optional videos# Proof of fundamental theorem of calculus

The first part of the fundamental theorem of calculus tells us that if we define 𝘍(𝘹) to be the definite integral of function ƒ from some constant 𝘢 to 𝘹, then 𝘍 is an antiderivative of ƒ. In other words, 𝘍'(𝘹)=ƒ(𝘹). See why this is so. Created by Sal Khan.

## Want to join the conversation?

- What is squeeze theorem?

I did read up, but I did not understand much of it....

<edited> I watched the video and I understand it now. I'm leaving this here (and not deleting it) just for the record.

Thanks for all the help everyone - particularly @TheCatofSauron !(22 votes)- I cannot explain this as well as the video can (It is in the limits section of calculus), but I will try anyway. Suppose I say that I have always been taller then or the sam height as my sister, but I've always been shorter than or the same height as my brother. If I then say that three months ago, my sister was 5' 11" and my brother was 5' 11", then my height must be in squeezed in between them, so my height was also 5' 11".

This helps in calculus if you know a variable, let's call it x, is less or equal to than y, but greater than or equal to z. Then if you find out that y and z are both equal to k, then x must be equal to k as well.

Of course, in calculus, it won't be that simple. It will involve sines and cosines and stuff, but I hope this helps! :)(65 votes)

- 5:45; Is there a video on the mean value theorem for definite integrals?(18 votes)
- At5:50, Sal uses the Mean Value Theorem as a tool to prove the Fundamental Theorem of Calculus (F.T.C); but, isn't the F.T.C needed to demonstrate the MVT for integrals? If not, could someone please show me a proof of it that doesn't use the FTC?

Thank you very much!(11 votes)- The rough proof is something like this:

let g(t) be a continuous function on [a, b]

By the extreme value theorem we can write m <= g(t) <= M.

Therefore we can write m*(b-a) <= integral from a to b of g(t) <= M*(b-a). (There is a smaller box that has area less equal to the area under g(t) which is less equal to the area of some bigger box)

Then we can write m <= (integral from a to b of g(t))/(b-a) <= M.

From the intermediate value theorem there is some c in [a, b] such that g(c) = (integral from a to b of g(t))/(b-a). Then we are done. (The intermediate value theorem says for [a, b] if f(a) <= S <= f(b) then there is a c in [a, b] such that f(a) <= f(c) = S <= f(b) of course if f is continuous.)

I guess the proof was kind of long and Sal did not have videos on the intermediate value theorem is why he didn't prove it. This is 4 months late but maybe it can help someone else.(12 votes)

- We know that derivatives graph the slope of a function. I wonder, what is the "slope" of an area, and how is it related to the function under which this area is determined?(9 votes)
- I think maybe it is clearer to think of
`F(x)`

as being a function that describes an area. Then`f(x)`

is the rate of change in the area (i.e. how much the area changes for an increment of`x`

).

In other words,`f(x)`

does not give the slope of the area, it gives the slope of the graph that describes how area changes with`x`

.(6 votes)

- When you take the derivative of a derivative (anti-derivative) using the Power Rule (not the reverse power rule), are you essentially applying the Fundamental Theorem of Calculus?(0 votes)
- The derivative of a derivative is not an anti-derivative.(28 votes)

- If I am not mistaken, the Fundamental Theorem of Calculus says that the indefinite integral of f(x) is the anti-derivative F(x). Can we say the reverse, that the derivative F'(x) is the anti-integral f(x)?

I hope I said this right. What I think that I am really asking is: is there a concept called "anti-integral" related to derivative similar to the relationship between integral and anti-derivative?

If you can figure out what I am trying to say, can you state it better?

Thanks for your help and insight.(4 votes)- The theorem states, leaving out all the details, that:

1. The derivative of the indefinite integral of f(x) is f(x).

2. The indefinite integral of the derivative of f(x) is f(x) + C.

Both versions of the statement require the function is continuous, real-valued and differentiable.(9 votes)

- where did the first definition come from .. i mean : how did we know that int(from a to x) f(x) = F(x) ?(3 votes)
- Sal literally just defined it that way. ∫ f(t) dt from a to x is the area under the f(t) graph. It is a function of x. The thought is instead of writing this cumbersome integral, why don't we just call it F(x) and save ourselves some effort?

Eventually, we see that this F(x) is the antiderivative of f(x).(8 votes)

- In order to prove this theorem to begin with, we had to have a pretty good hunch that F'(x) = f(x). What would have been the thought process of earlier mathematicians to lead them to believe that F'(x) = f(x)?(5 votes)
- It probably wasn't quite "out of the blue." I suspect Newton and Leibnitz would have realized that they could use the concept of infinitesimals, which they used to get to derivatives, to calculate area. Then, just trying a few functions, and plotting the areas as they varied the values of x (calculating area by brute force), it wouldn't take long to notice that the derivative of the area function was the original function.

Or, maybe one of them just noticed that*the derivative of pi r^2 = 2pi r*, and the rest is history.(3 votes)

- At10:32you say that c is a function of Δx.

What do you mean by this? Isn't c is a function of f(c)?(2 votes)- He means that c is determined by the ∆x because it always within the ∆x difference.

Also, "c is a function of f(c)" is a common error. Notice that c is in the parenthesis. What really happened is that f is a function of x – i.e. f(x) – and c is just a particular value of x.(6 votes)

- I was wondering.. Do I have to take single variable calculus course after taking this AP calculus ab? or are they just the same?(2 votes)
- It all depends - but normally a high score exempts (4 or 5) will exempt you from taking Calculus at a College/University. However, you can retake Single Variable Calculus at College/University and with good reasons. One of the first would be that many Universities teach an Honors Calculus course, which goes much deeper into the theory and logical structure of Calculus; or maybe you feel that you're not entirely ready to jump into Multivariable Calculus, or Linear Algebra. Or maybe you'd like a lighter first term at School and not stress yourself out as much with a lot of difficult courses along with physics, engineering, computer science, etc.(2 votes)

## Video transcript

Let's say that we've
got some function f that is continuous
on the interval a to b. So let's try to see if
we can visualize that. So this is my y-axis. This right over here, I
want to make it my t-axis. We'll use x little bit later. So I'll call this my t-axis. And then let's say that
this right over here is the graph of y
is equal to f of t. And we're saying it's continuous
on the interval from a to b. So this is t is equal to a. This is t is equal to b. So we're saying that
it is continuous over this whole interval. Now, for fun, let's define
a function capital F of x. And I will do it in blue. Let's define capital F of x as
equal to the definite integral from a is a lower
bound to x of f of t-- let me do that in--
of f of t dt, where x is in this interval, where
a is less than or equal to x is less than or equal to b. Or that's just
another way of saying that x is in this
interval right over here. Now, when you see this,
you might say, oh, the definite integral, this
has to do with differentiation an antiderivatives and all that. But we don't know that yet. All we know right
now is that this is the area under the curve f
between a and x, so between a, and let's say this
right over here is x. So f of x is just this
area right over here. That's all we know about it. We don't know it has anything
to do with antiderivatives just yet. That's what we're going to
try to prove in this video. So just for fun, let's
take the derivative of f. And we're going to
do it just using the definition of
derivatives and see what we get when we take
the derivative using the definition of derivatives. So the derivative, f prime
of x-- well this definition of derivatives, it's the
limit as delta x approaches 0 of capital F of x
plus delta x minus f of x, all of that over delta x. This is just the definition
of the derivative. Now, what is this equal to? Well, let me rewrite it using
these integrals right up here. This is going to be
equal to the limit as delta x approaches 0 of--
what's f of x plus delta x? Well, put x in right over here. You're going to get the definite
integral from a to x plus delta x of f of t dt. And then from
that, you are going to subtract this business,
f of x, which we've already written as the definite integral
from a to x of f of t dt, and then all of that
is over delta x. Now, what does this represent? Remember, we don't know anything
about definite integrals or somehow dealing
with something with an antiderivative
and all that. We just know this is
another way of saying the area under the curve f
between a and x plus delta x. So it's this entire
area right over here. So that's this part. We already know what
this blue stuff is. Let me do it in that
same shade of blue. So this blue stuff
right over here, this is equal to all
of this business. We've already shaded this in. It's equal to all of this
business right over here. So if you were to take all
of this green area, which is from a to x plus
delta x, and subtract out the blue area, which
is exactly what we're doing in the numerator
what are you're left with? Well, you're going to
be left with-- what color have I not used yet? Maybe I will use
this pink color. Well, no, I already used that. I'll use this purple color. You're going to be left with
this area right over here. So what's another
way of writing that? Well, another way of writing
this area right over here is the definite
integral between x and x plus delta x of f of t dt. So we can rewrite this entire
expression, the derivative of capital F of x--
this is capital F prime of x-- we can rewrite it
now as being equal to the limit as delta x approaches
0-- this I can write as 1 over delta x times
the numerator. And we already figured
out the numerator. The green area minus the blue
area is just the purple area, and another way of
denoting that area is this expression
right over here. So 1 over delta x times
the definite integral from x to x plus
delta x of f of t dt. Now, this expression
is interesting. This might look familiar
from the mean value theorem of definite integrals. The mean value theorem
of definite integrals tells us there exists
a c in the interval see where-- I'll
write it this way-- where a is less than or equal
to c, which is less than-- or actually, let
me make it clear. The interval that we care
about is between x and x plus delta x--
where x is less than or equal to c,
which is less than or equal to x plus delta x,
such that the function evaluated at c-- so let me draw this c. So there's a c
someplace over here-- so if I were take the
function evaluated at this c-- so that's f of c
right over here. So if I were to take
the function evaluated at this c, which
would essentially be the height of this line,
and I multiply times the base, this interval, if I multiply
it times the interval-- and this interval
is just delta x. x plus delta x minus
x is just delta x. So if we just multiply
the height times the base, this is going to be equal to
the area under the curve, which is the definite integral from x
to x plus delta x of f of t dt. This is what the mean value
theorem of integrals tells us. If f is a continuous
function, there exists a c in this interval
between our two end points where the function evaluated
at the c is essentially, you could view it
as the mean height. And if you take that mean
value of the function and you multiply
it times the base, you're going to get
the area of the curve. Or another way of
rewriting this, you could say that there exists
a c in that interval where f of c is equal to 1 over
delta x-- I'm just dividing both sides by delta x-- times
the definite integral from x to x plus delta x of f of t dt. And this is often
viewed as the mean value of the function
over the interval. Why is that? Well, this part right over
here gives you the area, and then you divide
the area by the base, and you get the mean height. Or another way you
could say it is, if you were to take the height
right over here, multiply it times the base, you
get a rectangle that has the exact same area as
the area under the curve. Well, this is
useful, because this is exactly what we got as the
derivative of f prime of x. So there must exist
a c such that f of c is equal to this stuff. Or we could say that the limit--
and let me rewrite all of this now in a new color. So there exists a
c in the interval x to x plus delta x where
f prime of x, which we know is equal to this, we can now
say is now equal to the limit as delta x approaches 0. And instead of
writing this, we know that there's some c that's
equal to all of this business, of f of c. Now we're in the home stretch. We just have to figure out
what the limit as delta x approaches 0 of f of c is. And the main realization is
this part right over here. We know that c is always
sandwiched in between x and x plus delta x. And intuitively, you
can tell that, look, as delta x approaches 0, as
this green line right over here moves more and more to
the left, as it approaches this blue line, the c
has to be in between, and so the c is
going to approach x. So we know intuitively
that c approaches x as delta x approaches 0. Or another way of
saying it is that f of c is going to approach f of
x as delta x approaches 0. And so intuitively,
we could say that this is going to be equal to f of x. Now, you might say,
OK, that's intuitively, but we're kind of working
on a little bit of a proof here, Sal. Let me know for sure that
x is going to approach c. Don't just do this little thing
where you drew this diagram and it makes sense
that c's going to have to get closer
and closer to x. And if you want
that, you could just resort to the squeeze theorem. And to resort to
the squeeze theorem, you just have to view c
as a function of delta x. And it really is. Depending on your delta x, c's
going to be further to the left or to the right, possibly. And so I can just
rewrite this expression as x is less than or equal to c
as a function of delta x, which is less than or equal
to x plus delta x. So now you see that c is always
sandwiched between x and x plus delta x. But what's the limit of x
as delta x approaches 0? Well, x isn't dependent
on delta x in any way, so this is just going
to be equal to x. What's the limit of x plus
delta x as delta x approaches 0? Well, as delta x
approaches 0, this is just going to be equal to x. So if this approaches
as delta x approaches 0, and it's less than
this function, and if this approaches x
as delta x approaches 0, and it's always
greater than this, then we know from the squeeze
theorem or the sandwich theorem that the limit as
delta x approaches 0 of c as a function
of delta x is going to be equal to x as well. It has to approach the same
thing that that and that is. It's sandwiched in between. And so that's a slight--
we resort to the sandwich theorem-- it's a little
bit more rigorous-- to get to this exact result. As delta x approaches
0, c approaches x. If c is approaching x, then f of
c is going to approach f of x. And then we essentially
have our proof. F is a continuous function. We defined capital
F in this way, and we were able to use just
the definition of the derivative to figure out that the
derivative of capital F of x is equal to f of x. And once again, why
is this a big deal? Well, it tells you that if you
have any continuous function f-- and that's what we assume. We assume that f is continuous
over the interval-- there exists some function--
you can just define the function
this way as the area under the curve between some
endpoint, or the beginning of the interval, and
sum x-- if you define a function in that way, the
derivative of this function is going to be equal to
your continuous function. Or another way of saying
it is that you always have an antiderivative, that
any continuous function has an antiderivative. And so it's a couple
of cool things. Any continuous function
has an antiderivative. It's going to be
that capital F of x. And this is why it's called
the fundamental theorem of calculus. It ties together
these two ideas. And you have
differential calculus. You have the idea
of a derivative. And then in integral
calculus, you have the idea of an integral. Before this proof, all
we viewed an integral as is the area under the curve. It was just literally
a notation to say the area under the curve. But now we've been able to
make a connection that there's a connection between the
integral and the derivative, or a connection between the
integral and the antiderivative in particular. So it connects all of calculus
together in a very, very, very powerful-- and we're
so used to it now, and now we can say almost
a somewhat obvious way, but it wasn't obvious. Remember, we always
think of integrals as somehow doing
an antiderivative, but it wasn't clear. If you just viewed an
integral as only an area, you would have to go
through this process to say, wow, no, it's connected
to the process of taking a derivative.