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# Proof of fundamental theorem of calculus

See why this surpising result is true. Created by Sal Khan.

Video transcript

Let's say that we've got some function f
that is continuous, continuous on the interval a to b, so let's try to
see if we can visualize that. So, this is my y axis, that's my y axis. This right over here. I want to make it my t axis. We'll, we'll use x a 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. Y is equal to f of t. And we are saying that it is 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 are saying that it is continuous. 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. >From a as a lower bound to x to x, of, f of t, of f of t, of f of t dt, where x is in this interval, where. 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,
you know, the definite integral just has to do with differentiation and
anti-derivatives and all that, but we don't know that yet. All we know right now. Is that this area under the curve f
between a and x. So between a and let's say this right over
here, this right over here is x. So f of x is just, is just. This area, this area right over here. That's all we know about it. We don't know if 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 gonna do it just using the
definition of derivatives and see what that what we
get when we take the derivative using the
definition of derivatives. So we would get, so the derivative, f
prime of x, well, this definition of derivatives,
it's a limit as delta x approaches 0 of capital F of x plus delta
x minus F of x, all of that. 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. This is going to be equal to the limit,
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 gonna 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. You are going to subtract. This business, f of x, which we've already
written as the integral from a, the deft integral from a to x of f of
t, dt. And then all of that is over delta x. All of this is over, all of this is over
delta x. Now what is this represent? Remember, we don't know anything about
definite integrals is somehow dealing with something with an
anti-derivative and all that. We just know this, this is another way of
saying the area under the curve f between a and x
plus delta x; so it's the area under the curve
f, between a and x plus delta x, x plus delta
x. So, it's this entire area. It's this entire area. Right over here. So that's this part. We already know what this blue stuff is. This blue stuff, in that same shade of
blue. So, this blue stuff right over here. This is equal to all of this business. We've all ready 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 this blue area which is
exactly what we are doing in the numerator, what are
you left with? Well, you're going to be left with. 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. You, 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, 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. The numerator, we have already figured out
that the numerator, the green area minus the blue area is just the
purple area. Which is an, an another way of solving is denoting that area is this expression
right over here. So 1 over delta x times a definite
integral from x to x plus delta x of f of t. 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
intervals tells us, so the mean, mean value theorem of
definite. Definite integrals. Definite integrals tells us, there exists,
there exists a c in, in the interval, so I
could, a c. 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 now care about is
between x and x + delta x, where x. Where x is less than or equal to c, which
is less than or equal to x + delta x. Such that, such that the function
evaluated at c, the function evaluated at c: so let me draw
that c. So there's a c some place over here. So if I were to take the function
evaluated at the c. So that's f of c right over here. So if I were to take the function
evaluated at c, which would essentially be the height
of this line, and I multiply it 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. Times the base, that this is gonna be
equal to the area under the curve. It's going to be equal to the area under
the curve, which is the definite integral from x to x plus delta
x, x to x plus delta x. F of t, 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
endpoints, between our two end points where the function of value of
the c is essentially, you can 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 f of c, 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 interval from x to x
plus delta x. 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, 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 exact a c, such that, such
that f of c is equal to this stuff, or we can say
that the limit. And let me re-write all of this in a new
color. So there exists, there exists a c in the interval x to x plus delta x. Where, 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 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 could
tell that, look, as delta x approaches 0 as this delta x as this
green line, as this green line right over here moves more
and more to the left. As it approaches this, as it approaches
this blue line, as it approaches this blue line the c has to be in between
so the c is going to approach x. So we know intuitively. We know intuitively c approaches x as
delta x, delta x approaches 0 or another way of
saying it is f of x is going to approach f of x as delta x, as delta x approaches
0. So intuitively we can say that this is
going to be equal to f of, this is going to be equal to f of
x. Now you might say, okay that's
intuitively, but we're kind of working on a little bit of a proof
here, Sal. Tell me, 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 is gonna have to get
closer and closer to x. And if you want that then you can 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, cs gonna 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 sandwich
in between x and x plus delta x. But whats the limit of x as delta x
approaches 0? Well x isn't dependent on delta x in
anyway so this is just going to be equal to x. Whats the limit of x plus delta x, as
delta x approaches 0? Well as delta 0 approaches 0 this is just
going to be equal to x. So if this approaches x, as delta x
approaches 0 and it's less than this function, and if this approaches x as
delta 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, is going to be
equal to x as well. It has to approach the same thing as that
and that is. It is sandwiched in between. And so that's the slight we we resort to
the sandwich theorem. It is a little 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 f in this way, 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, is equal to, 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 we assume, we assume that f is continuous over the interval, there exists
some function. There exists a function, you can just
define the function this way is the area under the curve, between, between some endpoint,
or the, 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 anti-derivative. That any continuous function has an
anti-derivative. And so it's a couple of cool things. Any continuous function has an
anti-derivative. It's gonna be that capital, its' 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 ideas, 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. Is, 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 anti-derivative. In particular so it connects all of
calculus together in a very, very, very powerful, and we're
so used to it now and, and now we can say almost in a somewhat obvious way but it
wasn't obvious. Remember, we always think of integrals as somehow doing an anti-derivative but it
wasn't clear. If you just viewed an integral as only an
area you would have to go through this process and say wow, no it's
connected, it's connected to the process of taking a
derivative.