Prove the Fundamental Theorem of Integral Calculus

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Lesson Parent: 
The Fundamental Theorem of Integral Calculus
Problem: 

Prove the Fundamental Theorem of Integral Calculus

Answer: 

It is true that for a function integrable on $[a,b]$ that $\int_a^bf(x)dx=F(b)-F(a)$ where $F(x)$ is an antiderivative of $f(x)$.

SAMPLE SOLUTION

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Solution: 

This is the most general proof of the Fundamental Theorem of Integral Calculus. If you are in a Calculus course for non-mathematics majors then you will not need to know this proof so feel free to skip it. If you are a math major then we recommend learning it. If you are some other science major then you should check with your professor on whether learning this proof will be required in your particular Calculus course.

The Fundamental Theorem of Integral Calculus states that if $f(x)$ is an integrable function on the closed interval $[a,b]$ and $F(x)$ is an antiderivative of $f(x)$, then

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Let's assume that we have partitioned the closed interval $[a,b]$ with a partition we denote by $P$.

Let's examine the sum

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Let's expand this out to see what it does:

<Sign in to see all the formulas>


(When a summation has this behavior that all of the "internal" values cancel we call it a telescoping series.) But recall that $a=x_0$ and $b=x_N$ in our notation for partitions. This means that

<Sign in to see all the formulas>

But our assumption in the problem is that $F(x)$ is an antiderivative of $f(x)$. This means that <Sign in to see all the formulas>. Now for each closed interval <Sign in to see all the formulas> in the partition, we can use the Mean-Value Theorem to find a point $e_i$ in each <Sign in to see all the formulas> such that

<Sign in to see all the formulas>

Now let's do a little algebra:

<Sign in to see all the formulas>


Now let's do the sum:

<Sign in to see all the formulas>


where $R(P)$ is the Riemann sum of $f(x)$.

It is extremely important to point out that $P$ is arbitrary. We did not specify a specific partition for $P$. This means that <Sign in to see all the formulas> is true for every partition. Keep this in mind for what follows.

The problem says that we are to assume that $f(x)$ is integrable. From our lesson on Riemann sums we know this means that for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that for every partition where <Sign in to see all the formulas> and any choice of <Sign in to see all the formulas> in each <Sign in to see all the formulas> then

<Sign in to see all the formulas>


Since this is true for any choice of <Sign in to see all the formulas> in each <Sign in to see all the formulas> then let's choose <Sign in to see all the formulas> so that <Sign in to see all the formulas>. (Remember we can do this because <Sign in to see all the formulas> is true for any partition.) So now for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that for every partition where <Sign in to see all the formulas> then

<Sign in to see all the formulas>

The only way this statement can be true is if <Sign in to see all the formulas>. If this were not true then there would be some number (let's call it <Sign in to see all the formulas>) where

<Sign in to see all the formulas>


But then for any <Sign in to see all the formulas> the statement

<Sign in to see all the formulas>


would not be true. But we are told to assume that it is! So the only we it can be true is if

<Sign in to see all the formulas>

The proof is now complete.