FTOC (Informal)

Let $f_0$ and $f_1$ be two functions such that $f_1$ represents the rate at which $f_0$ is changing with regards to some independent variable $t$. (If you prefer to read an example first, jump down to that section.)

This relationship is typically represented by saying that $f_1(t)$ is the derivative of $f_0(t)$, that is: $$f_1(t)=f_0^\prime(t)$$ or $$f_1(t)=\frac{\text{d}(f_0(t))}{\text{d}t}$$

This relationship is also represented by saying that $f_0(t)$ is the integral of $f_1(t)$, that is: $$\int f_1(t) \text{ d}t = f_0(t) + C$$ The $C$ captures the fact that the derivative of a constant is zero, and hence all functions equal to $f_0$ except for an added constant have the same derivative.

There are some restrictions on the functions $f_0$ and $f_1$, which amount to assuming that $f_0$ has a meaningful rate of change. Exploring those restrictions is a significant part of a calculus course.

The Fundamental Theorem of Calculus (FTOC) is, on one level, a formalization of the relationship between $f_0$ and $f_1$, that is, between a function and its derived function.

Part 1 says, in essence: If $\displaystyle{f_0(x) = \int_a^x f_1(t)\text{ d}t}$ then $f_0^\prime(x)=f_1(x)$.

Part 2 says, in essence: If $f_0^\prime(x)=f_1(x)$ then $\displaystyle{\int_a^b f_1(x)\text{ d}x = f_0(b) – f_0(a)}$.

A key discovery within calculus is that the integral represents the “accumulation” of the derived function, that is, that the integral of a function over a given interval is the area of the region between the graph and the $x$-axis.

The rest of calculus is exploring the ramifications of this relationship, as well as the situations in which it “breaks”.

Example

Let’s say that $f(x)=x^2$. Here’s the graph:

On the left side of the graph, it is very steep going downward. At the origin, it is flat. On the right side, it is very steep going upward. Hence, if we were to graph its rate of change at any value, we would expect to get a function that starts out very negative, moves to zero at the origin, and then becomes very positive.

One such function, which is indeed the derivative function, is $g(x)=2x$:

We can hence say that $$g(x)=f'(x)=2x$$ and that $$\int 2x \text{ d}x = x^2 + C$$

Furthermore, the region between the $x$-axis and the graph of $g(x)$ from $x=0$ to $x=5$ has an area of $$\int_0^5 2x \text{ d}x = x^2 \Bigm|_0^5 = 5^2 – 0^2 = 25$$ This can be confirmed by looking at the graph of $2x$ above.