The Fundamental Theorem of Calculus: Part 1

A definite integral $\displaystyle\int_a^b f(x)\,dx$ with both limits fixed is a single number. Suppose instead we hold the lower limit at $a$ and let the upper limit move. Each choice of upper limit gives its own number, so this defines a function of the upper limit.

The upper limit is now the input $x$, so the variable of integration inside needs a different letter, and we use $t$. It is a dummy variable: the choice of letter changes nothing about the value, it only keeps the two roles apart.

To evaluate $A$ at a particular input we integrate from the base point up to that input, so $A(3)$ is the net signed area from $a$ to $3$. At the base point itself the interval has zero width, so $A(a) = 0$.

Moving $x$ to the right adds more signed area to the running total, and each further piece takes its sign from whichever side of the axis $f$ is on.

The graph below shows a continuous function $f$. The shaded region above the axis has area $4$, and the shaded region below the axis has area $6$, split into two halves of area $3$ by the dashed line at $x = 4$.

For $A(x) = \displaystyle\int_0^x f(t)\,dt$, find $A(2)$, $A(4)$ and $A(6)$.

Solution

The base point is $0$, so each value of $A$ is the net signed area from $0$ up to the input. We take the three inputs in turn, keeping a running total.

For $A(2)$, the interval $[0, 2]$ covers exactly the region above the axis, which has area $4$. Everything here is above the axis, so it contributes positively:

For $A(4)$, we continue from $x = 2$ to $x = 4$. That interval adds the left half of the region below the axis, of area $3$. Below the axis the contribution is negative, so we subtract:

For $A(6)$, we add the right half of the same region, again of area $3$ and again negative:

Given a function $f$, we have been exploring two questions independently:

1. Find a function (the antiderivative) whose derivative is $f$, which we answered by reversing the rules of differentiation.
2. Find the net signed area beneath $f$, which we answered by adding up rectangles and taking a limit.

The two questions use different machinery and were built for different purposes. We do write $\int$ for antiderivatives as well as for areas, but nothing so far connects the two ideas; the shared symbol is a convention, not a result.

Astonishingly, they turn out to be the same question. That is the central result of calculus, the Fundamental Theorem of Calculus (FTC), found independently by Newton and by Leibniz in the second half of the seventeenth century. It comes in two halves; this lesson states the first.

The accumulation function, $A(x) = \displaystyle\int_a^x f(t)\,dt$, is what links the two. It is a signed area, and it is built out of $f$, so we can ask what its derivative is.

We can already read this much off the graph of $f$: $A$ increases exactly where $f$ is positive, decreases exactly where $f$ is negative, and is stationary exactly where $f$ is zero.

This is the behaviour of a function whose derivative is $f$. In other words, we claim $A' = f$.

The diagram below builds the case for that claim in four steps. Work through them with the buttons, dragging $x$ and adjusting $h$ as you go.

We state FTC Part 1 without proof.

So what does it mean? The equation is short, but the idea inside it is not, so it is worth saying plainly what the theorem claims.

Imagine painting a wall as you walk steadily to the right, the wall's height varying along its length. Let $A(x)$ be the amount of paint used by the time you reach position $x$. How fast are you getting through paint at this moment? It depends only on how tall the wall is where the brush is: in the next instant you cover a thin vertical strip, as wide as the small step you take and as tall as the wall there. With a tall wall the paint is disappearing quickly; with a low wall you're using barely any.

This is what $A' = f$ says. The height of $f$ at $x$ is the rate at which accumulated area is changing at $x$. Where the curve is high above the axis, area accumulates quickly and $A$ climbs steeply. Where the curve is close to the axis, almost nothing is being added and $A$ is nearly flat. Where the curve is below the axis, area is being taken away and $A$ falls. The height of $f$ is never the amount of area accumulated; it is the rate at which that amount changes.