Welcome to a Proof
of the Fundamental Theorem of Calculus, part one.
The Fundamental Theorem of Calculus part one states
that if a function f of x
is continuous on the closed interval from a to b,
and big f of x equals the definite integral of f of t,
with respect to t,
from a to x is continuous on the same closed interval
and differentiable on the open interval,
then big f prime of x,
which would be equal to the derivative with respect to x
of the definite integral of the f of t,
from a to x equals f of x, equals f of x.
Notice how this is telling us
the derivative of this integral
is equal to the integrand function f,
with the change of variables to x.
To begin our proof,
we'll use the definition of the derivative
for big F prime of x,
which equals the limit as h approaches zero
of the difference quotient for big F.
For the next step
we're going to factor out one over h from the fraction,
and then write big f of the quantity x plus h,
and big f of x as definite integrals using this equation here.
So the next step
we have the limit as h approaches zero of one over h,
and then times the quantity big f of the quantity x plus h,
is equal to this definite integral,
where we substitute the quantity x plus h for x,
and then we have minus big f of x
is equal to this definite integral here.
And now we'll apply properties of definite integrals.
So for the first step,
if we change the order of integration here
it's going to change the sign,
which would make this plus the integral of f of t
from x to a,
and then using the additive property of definite integrals,
the sum of these two integrals
would be equal to the integral of f of t
from x to the quantity x plus h.
To make sure we understand this,
let's look at this graphically.
When we do this, we'll assume f of t is not negative,
though it doesn't need to be.
Looking at the difference
of these two integrals graphically,
if this is the graph of f of t,
if we integrate f of t from a to the quantity x plus h,
we integrate along this interval here,
which again if f of t is not negative,
would give us this area here under the function.
Now if we subtract the integral of f of t from a to x,
we would be subtracting this area here,
and we could see would leave us the area
from x to a quantity x plus h,
giving us the defnite integral of f of t,
from x to the quantity x plus h.
Again we're illustrating this using a non-negative function,
but it's still true even if the function is negative.
So hopefully this helps explain
where we come up with this integral here.
And I highlighted this equation,
because we'll use it again later.
But now we're going to apply
the Mean Value Theorem for integrals,
so let's begin by reviewing this theorem.
The Mean Value Theorem for Integrals
states that if f of x is continuous
on the closed interval from a to b,
then there exists a number c
in the closed interval from a to b
such that the value of the definite integral is equal to f of c,
times the quantity b minus a.
Again to illustrate this,
we'll use a function f of x that's non-negative,
therefore we can think of this as area.
So if we integrate f of x from a to b,
where f of x is graphed in blue,
it will give us the area of the shaded region here,
and the Mean Value Theorem is telling us
that there is some value c on the closed interval,
let's say here, where f of c, this function value here,
which should be the height of this rectangle,
times the quantity b minus a,
which would be the width of the rectangle,
would give us the same value as the definite integral.
So if f of x is non-negative,
this means that the area under the function
would be the same as the area of this rectangle.
So now going back to our proof,
by the mean value theorem of integrals,
there exists a number c in the closed interval
from x to the quantity x plus h,
such that f of c times h is equal to the definite integral.
Remember h is the width of the interval.
So for this graph, c might be somewhere in here,
where f of c, the height of the rectangle, times the width h
would be equal to the value of the definite integral.
Now we multiply both sides by one over h,
we would have f of c equals one over h
times the definite integral.
Now as h approaches zero,
the quantity x plus h approaches x.
Again, let's look at this graphically.
So if this is h, notice as h approaches zero,
or moves in this direction here,
the quantity x plus h would have to approach x.
As a result as the quantity x plus h approaches x,
c would also have to approach x,
because c is in the closed interval
from x to the quantity x plus h.
Therefore we can say
the limit as h approaches zero of f of c,
would have to approach f of x,
and therefore this limit is equal to f of x.
Now we just need to put the pieces together.
Since we know that big f prime x is equal to the limit
as h approaches zero of one over h times this definite integral,
and we've also shown that f of c equals
one over h times the definite integral,
we now know that big f prime of x
would be equal to the limit of f of c,
which we know is equal to f of x, and we have our proof.
We now know that big f prime of x equals f of x.
And remember, big f prime of x was equal
to the derivative with respect to x
of the definite integral of f of t from a to x,
which we now know equals f of x.
So this is one possible proof
of the Fundamental Theorem of Calculus part one.
I hope you've found this helpful.