What the fundamental theorem of calculus connects

Slopes and areas look like two separate problems. What links the slope of a graph to the area under it?

Fix a start point a and let A(x) be the signed area under the graph of f from a to x. As x moves to the right, A changes by the signed area of each new strip, so the graph of f tells you how A moves. Stage 1 asks you to draw A from the graph of f alone.

Before you start · 3 quick checks on zoomed-in slopes, areas of simple shapes and signed area, about 3 min

Definitions

Signed area. For a function \(f\) and numbers \(p \lt q\), the integral \(\int_p^q f(t)\,dt\) is the signed area between the graph of \(f\) and the t-axis from \(p\) to \(q\): regions above the axis count as positive and regions below it as negative. When the region is made of rectangles, triangles and trapezoids, the integral is the sum of their areas, each with its sign.

Orientation. For the other order of the limits,

\[ \int_p^p f(t)\,dt = 0, \] \[ \int_q^p f(t)\,dt = -\int_p^q f(t)\,dt. \]

With this rule, additivity holds for any three points \(p\), \(q\), \(s\), in any order:

\[ \begin{aligned} \int_p^s f(t)\,dt &= \int_p^q f(t)\,dt \\ &\quad + \int_q^s f(t)\,dt. \end{aligned} \]

Area function. Fix a start point \(a\). The area function of \(f\) from \(a\) is

\[ A(x) = \int_a^x f(t)\,dt. \]

The letter \(t\) only names the variable of integration, so that \(x\) can be the upper limit. \(A(a) = 0\). For \(x \gt a\), \(A(x)\) is the signed area from \(a\) to \(x\); for \(x \lt a\), it is minus the signed area from \(x\) to \(a\).

Fine print: which functions have an integral

The integral is defined as the limit of sums of signed rectangle areas, \(f(t_i)\,\Delta t\), as the rectangles get thinner. On a closed interval, every continuous function has an integral, and so does every bounded function that is continuous except at finitely many jump points, such as a step function. Every function on this page is of one of these two kinds. Changing the value of \(f\) at a single point changes no integral, because a single point has no width.

1 See

Sketch the area function from the graph of f

Choose \(f\). The upper figure shows its graph for \(t\) from 0 to 4, and dashed vertical lines cut it into strips of width 0.5. The start point is \(a = 0\), so \(A(0) = 0\): the lower figure marks that point with a small square. Drag each of the eight handles up or down to where you think \(A(x) = \int_0^x f(t)\,dt\) is at \(x = 0.5, 1, \ldots, 4\). You can also focus a handle and use the arrow keys (steps of 0.05), Page Up and Page Down (0.5), Home (\(-1.5\)) and End (4.5). Then press Check my sketch. The page compares the change of your sketch over each strip with that strip's signed area; a strip counts as close when the two are within 0.05.

2 Try

Constants, start points and slopes

Goals: read the slope of A from f

The explorer below has the same five functions, with the actual area function \(A(x) = \int_0^x f(t)\,dt\) drawn. Drag the handle on the t-axis of the upper figure, or focus it and use the arrow keys (steps of 0.05). The upper figure shades the signed area from 0 to \(x\); the lower figure marks the point \((x, A(x))\) with a small square and draws a dashed segment through it with slope \(f(x)\).

3 Prove

Why the slope of A is f

Theorem (the fundamental theorem of calculus)

Assumptions. \(I\) is an open interval, \(f\colon I \to \mathbb{R}\) is continuous, \(a\) is in \(I\), and \(A(x) = \int_a^x f(t)\,dt\) for \(x\) in \(I\). A function \(F\) is an antiderivative of \(f\) on \(I\) when \(F'(x) = f(x)\) for every \(x\) in \(I\).

Part 1. \(A\) is differentiable at every \(x\) in \(I\), and

\[ A'(x) = f(x). \]

Part 2. If \(F\) is any antiderivative of \(f\) on \(I\), then for all \(p\) and \(q\) in \(I\),

\[ \int_p^q f(t)\,dt = F(q) - F(p). \]

The proof of part 1 uses additivity and the orientation rule from the definitions, the squeeze theorem, and two facts:

Extreme value theorem. A continuous function on a closed interval takes a smallest and a largest value there.
Rectangle bound. If \(p \lt q\) and \(m \le f(t) \le M\) for every \(t\) from \(p\) to \(q\), then \(m\,(q - p) \le \int_p^q f(t)\,dt \le M\,(q - p)\).

Proof of part 2, and why the constant of integration disappears

So any antiderivative will do, and the \(C\) in \(\int f(t)\,dt = F(t) + C\) never affects a definite integral. The difference \(F(q) - F(p)\) is often written \(\big[F(t)\big]_p^q\).

Part 1 is used on the page PMF, PDF and CDF: a CDF \(F(x) = \int_{-\infty}^x f(t)\,dt\) is an area function of the density \(f\), so \(F'(x) = f(x)\) wherever \(f\) is continuous. Each derivative here is the slope of a zoomed-in graph, as on the page What a derivative measures when you zoom in.

4 Apply

Problems

Each problem has hints and a full solution.