# Fundamental Theorem of Calculus, Part 1 (Explaining Calculus #16)

We have recently talked about two concepts that appear rather disparate. We have discussed antiderivatives – which is the idea of reversing all the rules for taking derivatives, and Riemann sums, which are used to calculate the areas underneath complicated shapes. What do these have to do with each other? It does not look like they are very related.

However, they are actually extremely close related – in fact these are essentially the same exact thing! The purpose of our discussion in this part of the series on calculus is to figure out why these two very different concepts are actually just two ways of talking about the same thing.

Functions that Calculate Areas

The first thing to introduce in order to make a connection between functions and areas. To do this, for a previously known function $f(x)$, the so-called area function $A(x)$. We will define $A(x)$ so that the value of $A(c)$ is the area between the $y$-axis, the vertical line $x = c$, the horizontal line $y = 0$, and the graph $y = f(x)$. See the blue shaded region below for an example (for now, ignore the red region, it will matter later).

The nice thing about $A(x)$ is that it enables us to discuss the idea of area (which is geometric) in the language of functions, which is somewhat foreign to geometry. Functions are very algebraic, and areas are very geometric. As usually happens when we find ways to mix together two different types of math, things become a little bit delicate and tricky, but we gain the benefit of being able to use the toolkits of both algebra and geometry when talking about $A(x)$. And now, since we know how to do calculus with functions, perhaps we can try to do some calculus with $A(x)$. This will be our goal.

In particular, we are going to want to learn how to take a derivative of $A(x)$. This involves studying the red sliver in the image above. Recall that the formula for the derivative of a function is

$A^\prime(x) = \lim\limits_{h \to 0} \dfrac{A(x+h) - A(x)}{h}.$

In order to understand what the value of this derivative might be, we need to understand to get a grip on the region that is included by $A(x+h)$ and $A(x)$. In order to do this, we will think about the problem from two different angles and put them together to see what we can learn.

Estimating a Sliver of Area (Method #1)

Our first method has essentially already been done. I have stated earlier in our discussion that the good thing about having an object that is both algebraic and geometric is that we can look at it from two different angles and learn different kinds of information from those two perspectives.

This first method is the algebraic method. By definition, $A(x)$ tells us about the area from $0$ all the way to the point $x$, and $A(x+h)$ pushes a little bit further to $x+h$. If we want to know what the red region is, the region beyond $x$ leading up to $x+h$, we can just substract the area counted in $A(x)$ from the slightly bigger area counted by $A(x+h)$. In the picture above, we could say

Red Area = (All Area to Left of $x+h$) – Blue Area = $A(x+h) - A(x)$.

This is one way we could think about calculating the red area.

Estimating a Sliver of Area (Method #2)

The second way we can think about counting up the red area goes back to geometry. Recall that when we wanted to talk about areas underneath complicated-looking curves, we employed the method called Riemann sums, whereby we calculated the area in two steps:

(1) Estimate the area using rectangles,

(2) Take a limit as the number of rectangles becomes infinitely large to make the formula exact.

Well, perhaps we could take a Riemann sum view and estimate the red area using a rectangle. But what would this rectangle be?

In order to calculate the area of a rectangle, we need two pieces of information – we need the length of the base and the height of the rectangle. The base stretches from $x$ from $x+h$, which is a base length of $(x+h) - x = h$. Now, the red region doesn’t exactly have one single “height” that we can use, since the top is curved, but the upper left corner is a convenient value to use since we actually know what it is – we know the height in the upper left hand corner is $f(x)$. Therefore, in order to get an initial estimate on the red area, we can take our height to be $f(x)$.

So, we can say that

Red Area $\approx f(x) * h$.

In line with the ideas that we use with Riemann sums, we could transform this $\approx$ to an equals sign by taking a limit as the width of our rectangle becomes infinitely small. If we did this limit right now, we couldn’t learn very much. The clever thing to do, now, will be to first mix in new information to the equation from our first method and only then take a limit as $h \to 0$.

Comparing the Two Methods

We have just finished discussing two methods for evaluating the red area. Putting the two formulas we have devised together, we can now see that

Red Area = $A(x+h) - A(x) \approx f(x) * h$.

Now, since the last two terms form a genuine mathematical formula, we can drop our reference to the red area:

$A(x+h) - A(x) \approx f(x) * h$.

Now, this is looking pretty good. We can divide both sides by $h$ to solve for $f(x)$:

$f(x) \approx \dfrac{A(x+h) - A(x)}{h}$.

Now, when we employed our second method, I mentioned that we could take a limit as $h \to 0$ (i.e. taking the limit as the rectangles in a Riemann sum becomes infinitely thin) in order to transform the $\approx$ into a genuine equals sign. Now is the time to finally do this. If we remember the definition of the derivative mentioned earlier in the post,

$f(x) = \lim\limits_{h \to 0} \dfrac{A(x+h) - A(x)}{h} = A^\prime(x)$.

Therefore, we have arrived now at a rather interesting formula, $f(x) = A^\prime(x)$. What are we to make of this new strange formula?

Importance of the Connection to the Derivative

Let’s pause to remember three things we have learned recently:

• Remember what $A(x)$ actually is. $A(x)$ is a function that knows how to calculate the area underneath the graph of the function $y = f(x)$.
• Remember that the only way we know so far to calculate the areas underneath fancy curves is to use the extremely complicated Riemann sum process.
• Remember that, from our previous discussion on antiderivatives, that $F(x)$ is an antiderivative of the function $f(x)$ if and only if $F^\prime(x) = f(x)$.

Let’s now put this information together. We have just learned that $A(x)$ satisfies the equation $A^\prime(x) = f(x)$. This means, because of the third bullet point, $A(x)$ is an antiderivative of $f(x)$. In our last discussion, we talked about how you can actually derive formulas for antiderivatives by taking advantage of rules we already know that work for derivatives. This means that we now have two ways to calculate $A(x)$: the complicated Riemann sum process and a much simpler algebraic process of reversing derivative formulas. True, it is not necessarily easy to reverse derivatives, but it is much, much more straightforward than trying to actually evaluate a Riemann sum! So we now actually have a way better method for calculating complicated areas – we can use antiderivatives to calculate areas!

There remains still one problem. There isn’t just one antiderivative of a function $f(x)$. The antiderivatives of the function $f(x)$ all look like $A(x) + C$ for random constant numbers $C$. So, if I calculate an antiderivative $F(x)$ of $f(x)$, how do I know whether I have calculated $A(x)$ or, say, $A(x) + 7$ or $A(x) - 3$? The answer is that we need to learn how to evaluate $A(0)$. Since $A(0) + C$ is always different for different values of $C$, we can tell the difference between all the possible options if we learn what $A(0)$ is.

But what is $A(0)$? Well, this is one of the places the geometry really helps us. Remember that $A(x)$ is the area underneath a graph from the point $0$ up to the point $x$. So we need to calculate the area between $0$ and $0$. But that isn’t even an area… that is a vertical line! Lines have no area, and so $A(0) = 0$. So, we now know what is called the First Part of the Fundamental Theorem of Calculus:

Fundamental Theorem of Calculus, Part 1: If $A(x)$ is a function that tells us the area underneath the graph of $f(x)$ from 0 up to the point $x$, and if $F(x)$ is an antiderivative of $f(x)$ with $F(0) = 0$, then

$A(x) = F(x)$.

In words, calculating areas and calculating antiderivatives are the same process.

Conclusion

Part 1 of the Fundamental Theorem of Calculus is hugely important. It is called fundamental for a reason – it is upon the discovery of this massively important fact that calculus really begins to become supremely useful.

Next time, we will discuss part 2 of the fundamental theorem. Up to this point, the fundamental theorem has remained a bit abstract, in the sense that we are still dealing only with variables, and we have required that 0 be our starting point. Part 2 of the fundamental theorem removes these restrictions and enables us to place an actual number on areas underneath complicated curves.

Reference for Image

[1] By Kabel – Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=11034713