Welcome to a proof
of the Fundamental Theorem of Calculus, part 2.
In this proof, we'll be using the Mean Value Theorem.
The Fundamental Theorem of Calculus, part 2
states that if a function, f of x, is continuous
on the closed interval from a to b,
and big f of x is the antiderivative of f of x
on the closed interval,
then the def integral of f of x,
with respect to x from a to b
equals big f of b minus big f of a.
This tells us we can determine
the value of this def integral
by first determining the antiderivative big f of x,
then evaluating the antiderivative
at the upper limit of integration, which is b,
then at the lower limit of integration, which is a,
then determining the difference.
Now let's work on our proof.
For the proof, our goal is going to be
to write big f of b minus big f of a in a form
so that we can apply the Mean Value Theorem.
Consider the closed interval from a to b,
with the following partitions.
We partition the closed interval from a to b,
by letting a equal x sub zero and b equal x sub n,
and then we use x sub one
through x sub n minus one
to form the partitions,
which would look like this along a horizontal axis.
Now that we have our partitions,
it follows that to find big f of b minus big f of a,
we can sum increments of the function values
using the partitions, where big f of b minus big f of a
would be equal to the sum of these differences.
Let's look at this graphically
to make sure we understand this.
If we look at the graph of possible big f of x,
graphed here in blue, over the closed interval from a to b,
notice this value here would be big f of b,
and this value here
would be big f of a,
so big f of b minus big f of a
can be viewed as this length here,
and we're saying this distance here
would be the sum of the vertical distances
over each partition.
Looking at the first partition
from x sub zero to x sub one,
this vertical distance here
would be big f of x sub one minus big f of x sub zero.
Looking at the next partition,
this vertical distance here,
would be big f of x sub two minus big f of x sub one.
If we sum these vertical distances over all the partitions,
it would be equal to big f of b minus big f of a,
which is what we're saying here.
Now we're going to write this sum using sigma notation,
so we'd have big f of b minus big f of a
equals the summation of big f of x sub i
minus big f of x sub i minus one
from i equals one to n.
This notation here
would give us the same sum of these differences.
Now, but the Mean Value Theorem,
there exists a number, c sub i, in the ith subinterval,
such that big f prime of c sub i
is equal to the quotient of these two differences,
which, remember would give us
the slope of a secant line over one partition.
This is telling is there's one x value
within each subinterval
where the slope of the tangent line,
big f prime of c sub i,
would be equal to the slope of the secant line,
passing through the two endpoints of the closed interval.
Just in case you need it,
here's a quick review of the Mean Value Theorem.
If the conditions above here are met,
the Mean Value Theorem states
that f prime of c is equal to this quotient here,
where, again, this quotient is the slope of the secant line,
passing through the two endpoints of the closed interval,
and c is an x value in the open interval,
where the slope of the tangent line at that location
would be the same as the slope of the secant line.
Now we'll let delta x sub i be equal
to x sub i minus x sub i minus one,
and since big f prime of c sub i equals little f of c sub i,
because remember, big f of x
is the antiderivative of little f of x,
and therefore the derivative of big f equals little f.
Now we'll substitute delta x sub i
for x sub i minus x sub i minus one,
and we'll substitute little f of c sub i
for big f prime of c sub i,
which gives us this equation here.
Now, if we multiply both sides by delta x sub i,
and flip the equation around,
we would have big f of x sub i
minus big f of x sub i minus one
equals f of c sub i times delta x sub i.
This equation tells us
this difference is equal to this product
for each subinterval,
but if we want an equation
over the entire interval from a to b,
we would have big f of b minus big f of a
is equal to the sum of these products
from i equals one to n.
We need to recognize that as delta x sub i approaches zero,
meaning the width of the subintervals approach zero,
and the number of partitions, or subintervals,
would approach infinity.
Now we'll take the limit
of both sides of this equation here,
as it approaches inifinity.
The limit as it approaches infinity of the left
must equal the limit as it approaches infinity of the right.
Notice on the left, big f of b minus big f of a
is just going to be a constant,
and this is not affected by n,
and therefore on the left,
this is just going to be equal to
big f of b minus big f of a.
Now, looking on the right, we should recognize this limit.
This limit is the definition of a def integral.
For a quick review,
the definition of a def integral is given here,
and notice how this limit on the right
is the same limit we have in our proof.
Therefore, by definition of a def integral,
we have big f of b minus big f of a equals
the def integral of f of x,
with respect to x from a to b,
and we have our proof.
If change over this equation,
we have the Fundamental Theorem of Calculus, part 2.
The integral of f of x,
with respect to x from a to b,
equals big f of b minus big f of a.
I hope you found this helpful.