The Fundamental Theorem of Calculus

The Average Change of an Antiderivative

Let's look at FTC another way:

All we've done here is reverse the formula. Now, recall the relationship between ${F'}$ and FTC:

Keeping this in mind, let's rewrite our reversed formula using the following notation:

This yields:

$$\Delta F = \int_{a}^{b} f(x) dx$$

It turns out that this is the standard way of expressing FTC notationally. Say we divide the lefthand side by ${\Delta x.}$ We get:

The righthand side is a fairly important expression:

average change of an antiderivative. Let ${f}$ be a differentiable function. Then the average change of ${F}$ (the antiderivative of ${f}$), denoted ${\texttt{Avg}(F),}$ is:

$$\texttt{Avg}(F) = \dfrac{1}{b-a} \int_{a}^{b} f(x) dx$$

Just to be clear, the average of value of a function is the sum of the values of the function given an input ${x_0}$ to an input ${x_n,}$ divided by the number of inputs ${n.}$ That is:

This is a special type of Riemann sum:

That is, the integral from ${0}$ to ${n}$ where the increment is 1. If the increment were instead ${\lim\limits_{\Delta x \to 0},}$ we would get the continuous average of ${f.}$ With ${\Delta x = 1,}$ we have the discrete average of ${f.}$

The lemma we stated above tells us that the average change of the antiderivative of ${f}$ on ${a}$ to ${b}$ can be found by computing:

Using this fact, we can rewrite FTC formula again:

That is, the change in ${F}$ is the average infinitesimal change, times the total amount of change.

The Mean Value Theorem and The Fundamental Theorem of Calculus

Recall that the mean value theorem says:

Because ${F}$ is a function, the mean value theorem also applies:

Rewriting without substitute notation:

Comparing the two formulas, FTC formula is much more specific than the MVT. With MVT, the value ${c}$ likes somewhere between ${a}$ and ${b.}$ That is,

Accordingly, all we can conclude with the MVT is:

What this tells us is that FTC is a much strong conclusion that the MVT. It's obvious that the average is less than or equal to the maximum ${F'}$. And it's also obvious that the average is greater than or equal to the minimum ${F'.}$ Thus, we can draw the same conclusion that we would with the MVT:

The difference, however, is that ${\Delta F}$ is a very specific value:

That's far more specific and affirmative compared to what the MVT gives us: "Oh, it's ${F'(c)}$ times ${\Delta x}$ where ${c}$ is between ${b}$ and ${a.}$ What ${c}$ is, I couldn't tell you."

To illustrate, suppose we were given the following problem:

problem. Given the facts:

1. ${F(x) = \dfrac{1}{1+x}}$
2. ${F(0) = 1}$

For what values of ${a}$ and ${b}$ is the following proposition true:

$$a < F(4) < b$$

Using just the MVT, we have:

Thus, the range of possible values is ${4}$ to ${\dfrac{4}{5}.}$ It follows that:

And since ${F(0) = 1,}$ we have:

Now let's compare that to using FTC. With FTC, we have:

The largest possible value of ${F(4)-F(0)}$ is when ${x=0:}$

Since ${\int_{0}^{4} dx}$ is just the integral of ${1}$ from ${0}$ to ${4}$, we have:

The smallest possible value of ${F(4)-F(0)}$ is when ${x=4:}$

We know that ${\int_{0}^{4} dx = 4,}$ so we can deduce that:

So, we conclude that:

And, once again, using the fact that ${F(0) = 1,}$ we have:

Finally, because the integral is a summation of infinitesimals, we know that the inequalities are actually strict:

Restatement of the Fundamental Theorem of Calculus

Here is another statement of FTC:

restatement of the fundamental theorem of calculus

Iff:

1. ${f}$ is a continuous function.
2. ${G(x) = \int_{a}^{x} f(t)~dt.}$
3. ${a \leq t \leq x.}$

then:

$$G'(x) = f(x)$$

Or, alternatively:

$$\dfrac{d}{dx} \int_{a}^{x} f(t)~dt = f(x)$$

Let's be very clear about this proposition's components. First, the integrand's bounds are from ${a}$ to ${x.}$ ${a}$ is the lower bound, and ${x}$ is the upper bound. Thus, the area we're interested lies somewhere between, or is the entire area, beneath ${F(x) - F(a).}$

Second, we're using a dummy variable ${t.}$ The value of this variable lies anywhere between fixed values of ${a}$ and ${x.}$ Because it lies between these points, it's called the variable of integration. Do not, under any circumstances, mix ${x}$ and ${t.}$ This is a very common practice in older calculus textbooks, but it is extremely careless and dangerous.

The statement above is just an alternative statement of FTC. Why? Because if we're given an ${a,}$ and we're given an ${x,}$ we get the exact same statement we saw earlier. To see why this restatement is useful, let's just stare at the restatement's conclusion for a moment:

On its own, ${G'(x)}$ is a solution to the differential equation:

That is, ${G'(x)}$ is a solution to the differential equation ${y' = f,}$ where the initial condition is ${y(a) = 0.}$ The restatement tells us that we can always — always — solve the differential equation above. This is a foundational premise for the field of differential equations.

For example, consider this problem:

problem. Evaluate:

$$\dfrac{d}{dx} \int_{1}^{x} \dfrac{dt}{t^2}$$

On its face, it doesn't look like what we've discussed so far allows us to solve this. But, remember the restatement. The term:

is just ${G(x).}$ Thus:

And we know that:

What's ${f(x)?}$ It's the integrand:

Thus, we have:

Just to make sure we've done this correctly, let's actually integrate:

Integrating, we have:

Now, we know that:

If we differentiate ${G,}$ we get:

FTC Proof 1

Here is the first proof of FTC we'll present. Suppose we have the following graph:

Now say the area we're interested in is from ${a}$ to ${x,}$ as presented below:

Viewing the area ${\Delta G}$ as if it were a rectangle, we know that the base is ${\Delta x.}$ So, we now want to find its height. Well, the height is either ${f(x),}$ or ${f(x+\Delta x).}$ Ultimately, it doesn't matter. We know that ${f}$ is continuous (by assumption):

FTC Proof 2

Now let's examine the second proof. First, we start with the hypothesis:

hypothesis. ${F' = f.}$

Next, we will assume that ${f}$ is continuous.¹ Next, we will say that the following definition applies:

definition. ${G(x) = \int_{a}^{x} f(t)~dt}$

Now we apply our conclusion from the previous proof. There, we concluded that:

We also know that ${F'(x) = f(x)}$ (the derivative of the antiderivative of ${f}$ is ${f}$). Hence, it follows that:

And by the mean value theorem, we know that if two functions have the same derivative, they differ by a constant. Or, if a function has the derivative zero, the function is a constant. And since ${F(x)}$ and ${G(x)}$ have the same derivative, namely ${f,}$ it follows that:

where ${C}$ is some constant. All that's left to do is arithmetic:

We know that this integral runs from ${a}$ to ${b}$. Thus, the term ${G(a)}$ is zero, since it returns the integral from ${a}$ to ${a:}$

And ${G(b)}$ is just the integral of ${f}$ from ${a}$ to ${b:}$

FTC and Transcendentals

If the previous sections weren't enough to warrant FTC's namesake, perhaps the next sections will. It turns out that FTC opens an entirely new world of numbers and functions: transcendentals. These are numbers and functions outside the world of classical algebra. That is, nothing we've learned before FTC allows us to understand or work with objects in this realm. To better understand what this means, we begin by considering what FTC can tell us about functions we've already seen.

FTC Interpretation of Logarithmic Functions

The FTC provides an alternative way of defining and interpreting logarithmic functions. To begin, suppose we have the following derivative:

$$y' = \dfrac{1}{x}$$

By FTC, we can say that the logarithm is defined as:

$$L(x) = \int_{1}^{x} \dfrac{1}{t} ~dt$$

We know that:

$$L'(x) = \dfrac{1}{x}$$

If we evaluate the function ${L}$ at ${x=1,}$ we have:

$$L(1) = \int_{1}^{1} \dfrac{1}{t}~dt = 0$$

Next, let's compute the second derivative of ${L:}$

$$L''(x) = -\dfrac{1}{x^2}$$

Accordingly, where ${y' = \dfrac{1}{x},}$ we have the following conclusions and subconclusions:

1. ${L'(x) = \dfrac{1}{x}}$
   • At ${x=1,}$ ${L'(x) = 1.}$
2. ${L''(x) = \dfrac{1}{x^2}}$
   • ${L''}$ is concave down at every place.
3. ${L(1) = 0}$

With these premises, we can sketch a graph:

The actual graph:

Notice that from the sketch, we see see that when ${x < 1,}$ we have ${x \leq 1.}$ This behavior is expected. ${L(1) = 0,}$ and ${L}$ is increasing. This means that before 0, ${L}$ must be negative.

So, we see that FTC presents another way to define — and as a side-effect, sketch — the logarithmic function. Is there anything else? Sure. The FTC gives us another way to prove that:

$$L(ab) = L(a) + L(b)$$

where ${L}$ is a logarithmic function. The first step we can take is to just plug-and-play. To do so, we break the integral down into two parts:

$$\begin{aligned} L(ab) &= L(a) + L(b) \\ &= \int_{1}^{ab} \dfrac{dt}{t} \\ &= \int_{1}^{a} \dfrac{dt}{t} + \int_{a}^{ab} \dfrac{dt}{t} \end{aligned}$$

Now we're going to perform a little trick. We're going to say that:

$$t = au$$

Using this variable, we can say that:

$$dt = adu$$

And if we can say that, we can write:

$$\begin{aligned} \int_{a}^{ab} \dfrac{dt}{t} &= \int \dfrac{adu}{au} \\ \end{aligned}$$

Now we have to set the limits. First, we know that when ${t = a,}$ that's the lower limit. In that case, we have:

$$u = 1$$

Knowing this lower limit, we can add the lower bound to our rewritten integral:

$$\begin{aligned} \int_{t=a}^{ab} \dfrac{dt}{t} &= \int_{u=1} \dfrac{adu}{au} \\ \end{aligned}$$

Likewise, when ${t = ab,}$ we know that ${u = b.}$ That gives us the upper bound:

$$\begin{aligned} \int_{t=a}^{t=ab} \dfrac{dt}{t} &= \int_{u=1}^{u=b} \dfrac{adu}{au} \\ \end{aligned}$$

Cancelling out the ${a}$ in our rewritten integral, we get:

$$\begin{aligned} \int_{t=a}^{t=ab} \dfrac{dt}{t} &= \int_{u=1}^{u=b} \dfrac{du}{u} \\ \end{aligned}$$

which is precisely what we wanted to find.