Fundamental Theorem of Calculus, Part 2 (Explaining Calculus #17)

We have now discussed an extremely important tenet in the development of calculus, which is called the first part of the Fundamental Theorem of Calculus. We briefly recall what we deduced in the last post:

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

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

Informally, this told us that the process of calculating areas underneath graphs and the process of finding antiderivatives are really the same process. This is rather surprising at first – derivatives were originally designed to tell us about how things change over time, so it isn’t exactly obvious how reversing that process tells us about areas underneath graphs. There is, however, one example which I find particularly enlightening as to why this is the case. We will now explore this example and afterwards develop some new notation to more conveniently express both the first and second parts of the Fundamental Theorem.

An Example from Travel

Consider for a moment the following question:

Suppose you are running at a speed of 8 miles per hour for 5 hours. How far have you run?

A moments reflection reveals the answer is 40 miles. How do we know this? Well, we are taught in school that

Distance = Speed x Time.

We are told the speed, which is 8 miles per hour, and the time, which is 5 hours, and so we multiply them to see that the answer must be $8 \times 5 = 40$ . Now, let us reconsider this formula in terms of a graph. If you draw a graph with speed on the $y$ -axis and time on the $x$ -axis, then the running scenario we just discussed looks like a rectangle with height equal to 8 (the speed) and length equal to 5 (the time). The area of a rectangle is

Area = Length x Width.

This looks a lot like the above formula for distance – in fact, in this case they are the same formula and give the same answer of 40 miles. So we see where the area comes from. But the antiderivative is also quite present in this problem – the idea of speed is that speed tells you how quickly your location is changing. But if we remember that a derivative tells us how quickly something is changing, we should now understand that speed must be the derivative of the “location” function. So, if areas are the same thing as antiderivatives, then the area underneath a speed graph should tell us how much our location has changed. And this is correct – the area indeed tells us how far we have run.

This example should give the basic idea of why we might expect areas and antiderivatives to be connected. This example also hints at an additional way to understand antiderivatives as so-called “net changes” – that is, an antiderivative tells us something about how much some quantity has changed from a “start” time to some “end” time. In the example above, the area of 40 tells us that we have travelled 40 miles between when we started running and when we stopped running. This additional perspective, once understood, suggests that we probably have more to discover. This is true, but in order to write this down conveniently we need some new ways of writing down this type of information.

The Integral Notation

Up to this point, every time we wanted to talk about an antiderivative of a function $f(x)$ , we just had to write a new function $F(x)$ and declare that $F'(x) = f(x)$ . We will now introduce a different way of writing down this fact.

Indefinite Integrals

This new notation for writing down antiderivatives goes under the name of integral. So, talking about the integral of a function is the same thing as talking about the antiderivative of the function. The way we write down integrals is

$\int f(x) dx.$

The $dx$ is present in this formula to inform us that $x$ is the correct variable. So, for example, the function $\frac{x^3}{3} + C$ is the most general antiderivative of $x^2$ , and therefore we now write

$\int x^2 dx = \dfrac{x^3}{3} + C.$

These integrals are called indefinite integrals, or sometimes just integrals, because there is a second way the symbol $\int$ is used

Definite Integrals

One of the reasons we introduce this new symbol $\int$ is to define the definite integral. This is connected to the idea of antiderivatives, but is actually just a single number rather than a function. The so-called definite integral of a function $f(x)$ from $a$ to $b$ is written

$\int_a^b f(x) dx.$

What is this number? The value of $\int_a^b f(x) dx$ is the area underneath the graph of $f(x)$ between the value $x = a$ and $x = b$ .

For example, consider

$\int_0^3 2x dx.$

The graph of $y = 2x$ is a straight line. When $x = 0$ , we find that $y = 0$ , and similarly when $x = 3$ we find $y = 6$. When you sketch the area underneath this graph, you get a triangle. The vertices of this triangle are $(0,0), (3,0)$ , and $(3,6)$ . Its area is therefore

$\dfrac{1}{2} \mathrm{Base} \times \mathrm{Height} = \dfrac{1}{2} \left( 3 \times 6 \right) = \dfrac{18}{2} = 9.$

I encourage the reader to sketch this out for themselves.

Second Part of the Fundamental Theorem

We can now discuss the second part of the Fundamental Theorem of Calculus. This part really tells us how to actually calculate definite integrals in terms of antiderivatives – in other words, it tells us that definite and indefinite integrals are sort of two sides of the same coin, and that once again areas and antiderivatives are inseparable from one another.

Fundamental Theorem of Calculus, Part 2: Let $F(x)$ be an antiderivative of the function $f(x)$ , so $F'(x) = f(x)$ . Suppose $a,b$ are numbers with $a < b$ . Then

$\int_a^b f(x) dx = F(b) - F(a).$

Before we show the proof of this, let’s go back to the triangle example from earlier and show how we can compute the area of that triangle in this way. Since the integral of $2x$ is $x^2 + C$ for some unknown constant $C$ , we should have

$\int_0^3 2x dx = \left[ x^2 + C \right]_0^3 = \left( 3^2 + C \right) - \left( 0^2 + C \right) = 9 + C - C = 9.$

Notice that the $+C$ didn’t contribute at all to the final solution. For this reason, when using the fundamental theorem of calculus we will very often just set $C = 0$ since it won’t matter what its value is. Also note that this calculation gave the correct answer of 9 for the value of the area of the triangle.

Now, let us see why this way of calculating areas works.

Proof: Define the function

$A(x) = \int_0^x f(u) du.$

Remember that because of Part 1, we know $A(x)$ is the formula for the area underneath the graph of $f(x)$ between $0$ and $x$ and also the formula for the antiderivative $F(x)$ of $f(x)$ .

We will first focus on the fact that $A(x)$ counts areas. If we plug in $b$ , then $A(b)$ is the area underneath the graph of $f(x)$ between the $x = 0$ and $x = b$ , and $A(a)$ counts between $x = 0$ and $x = a$ . By subtracting out areas, the area underneath the graph of $f(x)$ between $x = a$ and $x = b$ is $A(b) - A(a)$ . But since $A(x) = F(x)$ , this area is also just $F(b) - F(a)$ , which is what we claimed it would be.

Conclusion

This finished our discussion of the fundamental theorem of calculus. We have now reached one of the peaks of the whole of calculus – we’ve discovered that the processes of calculating speeds and changes over time is the reverse of the process of counting areas, and we’ve learned how to ‘go back and forth’ between them. Now, since integrals are a relatively new introduction, we will next take some time to practice working with them and learning some useful tricks for dealing with more complicated integrals.

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