From derivatives to integrals

The running total

Fix a start $a$ and let $x$ slide. For a continuous function $f$, let $A(x)$ be its signed area from $a$ up to $x$: the accumulated total. How quickly does this total change as $x$ moves?

Slide $x$ right by $\Delta x$. The area grows by one new sliver, width $\Delta x$ and height $f(x)$:

$$A(x + \Delta x) - A(x) \;\approx\; f(x)\,\Delta x$$

Divide by $\Delta x$ and shrink:

$$A'(x) = f(x)$$

The slope of the running total is the height you are adding right now. Where the curve is tall, area piles up fast and $A$ climbs steeply. Where the curve is near zero, $A$ goes flat. For a continuous density, the running total of probability works this way: the density is the slope of the CDF.

The hill f(x) = x(3 − x) on top, its running total A below. The area under f between two points equals A's climb. A's slope at any point equals f's height.

Try it

Drag $a$ and $b$: the shaded area and the climb are always the same number. Switch to slope = height and drag $x$ to the hilltop at 1.5, where $A$ is steepest.

The shortcut

You want $\int_a^b f(x)\,dx$. Find any function $F$ whose slope is $f$. That is an antiderivative. On the interval from $a$ to $b$, $A$ and $F$ have the same slope, so they are the same curve shifted up or down: $A(x) = F(x) + C$. At $x = a$ nothing has been collected, so $A(a) = 0$, which means $C = -F(a)$. Therefore:

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

Find a function whose slope is the curve. Subtract its values at the two ends. For $x^2$, the slope of $x^3/3$ is $x^2$, so

$$\int_0^1 x^2\,dx \;=\; \frac{1^3}{3} - \frac{0^3}{3} \;=\; \frac13$$

This gives the same limit as the rectangle sums. For the hill in the widget, $f(x) = 3x - x^2$ has antiderivative $\tfrac32 x^2 - \tfrac13 x^3$, so the area from 0 to 3 is $\tfrac{27}{2} - 9 = 4.5$.

Small changes add up to the total change

Here is the same theorem read the other way. Take any smooth $F$ and chop $[a, b]$ into steps. The total change is the sum of the step changes. Every middle value appears once with a plus and once with a minus, and cancels:

$$[F(x_1) - F(a)] + [F(x_2) - F(x_1)] + [F(b) - F(x_2)] \;=\; F(b) - F(a)$$

Each step change is about $F'(x)\,\Delta x$, one sliver under the curve $F'$. Shrink the steps:

$$\int_a^b F'(x)\,dx \;=\; F(b) - F(a)$$

Add up all the small changes and you get the total change. Integrate speed, get distance. Integrate a slope, get the climb. This is the other direction of the fundamental theorem: integrating a continuous derivative recovers the total change.

Why the constant C appears

On an interval, any two antiderivatives of $f$ differ by a constant. $x^3/3$, $x^3/3 + 7$, and $x^3/3 - 100$ all have slope $x^2$. Written without ends, $\int x^2\,dx = x^3/3 + C$. The $C$ cancels in $F(b) - F(a)$. So antiderivatives are the derivative table read backwards:

Apply these formulas on an interval where the function is continuous. For arbitrary real powers, use $x>0$; for $1/x$, the interval cannot cross zero.

height $f$	an antiderivative $F$	because
$x^n$ ($n \neq -1$)	$\dfrac{x^{n+1}}{n+1}$	$\left(\dfrac{x^{n+1}}{n+1}\right)' = x^n$
$1/x$ ($x\ne0$)	$\ln\lvert x\rvert$	the log’s slope
$e^x$	$e^x$	its own slope
$c$	$cx$	a line’s slope is $c$

Some continuous functions have no antiderivative expressible using elementary functions such as polynomials, exponentials, and logarithms. One example is $e^{-x^2}$. Its integrals can still be defined and approximated numerically. The standard normal CDF $\Phi$ is a related integral, using the normalized density $e^{-x^2/2}/\sqrt{2\pi}$.

Go deeper: the telescoping sum, with numbers

$F(x) = x^2$ from 0 to 2 in four steps of $\tfrac12$. The step changes are $0.25, 0.75, 1.25, 1.75$, which sum to $4 = F(2) - F(0)$. The predictions $F'(x)\,\Delta x = 2x \cdot \tfrac12$ at each step’s left end are $0, 0.5, 1.0, 1.5$, sum 3. At the right ends, $0.5, 1, 1.5, 2$, sum 5. Shrink the steps and both close on 4.

Next

A model’s loss can depend on many parameters. The next lesson holds all but one input fixed to define partial derivatives, then combines them to predict changes in several inputs.