The expression

$$\int_0^2 x^2\,dx$$

has five pieces: the sign $\int$, the numbers $0$ and $2$, the function $x^2$ and the $dx$. Each does a different job, and each came from somewhere. Chapter 1 told the passage from Archimedes’ exhaustion to the integral. Here the integral is read slowly, the way one reads a sentence in a new language.

A sentence in five pieces

  1. The $\int$ is an elongated S, for summa, sum. Leibniz first wrote it in a manuscript dated 29 October 1675. It is an iconic sign in a curious way: it does not look like an area, but like the word that describes how the area is made.
  2. The limits $0$ and $2$, below and above, say from where to where one sums. They are not Leibniz’s: it was the Frenchman Joseph Fourier who, around 1820, began writing them attached to the sign. Before that, they were written out in words, alongside.
  3. The function $x^2$ is the height of each strip.
  4. The $dx$ is the width of each strip, infinitely thin. And it is also the marker of which letter varies: if it said $dt$, the variable would be $t$.
  5. The letter $x$ is a dummy letter: $\int_0^2x^2\,dx$ and $\int_0^2t^2\,dt$ have the same value. The result, $\frac83$, does not depend on $x$.

Read as a whole, the sentence says: add up the heights $x^2$ times the widths $dx$, for all $x$ from $0$ to $2$.

The sum behind the sign

This reading is not a metaphor. Replace the infinitely thin width $dx$ with a real width $\Delta x$, and the $\int$ with the Greek $\Sigma$, for sum. The integral becomes a sum of rectangles, the Riemann sum:

$$\sum_{i=1}^{n}f(x_i)\,\Delta x\;\longrightarrow\;\int_a^b f(x)\,dx.$$

The notation does here what it did in text 2: the shape stays as it passes to the limit. $\Sigma$ becomes $\int$, $\Delta x$ becomes $dx$, $f(x_i)$ becomes $f(x)$. Whoever understands the finite sum reads the integral effortlessly.

The forgotten dx

Many students write $\int x^2$, without the $dx$, and the teacher corrects it. It looks like pedantry, and it is not. The $dx$ does real work in a change of variable. Calculate

$$\int 2x\cos(x^2)\,dx.$$

With $u=x^2$, the differential is $du=2x\,dx$. The whole expression is transformed, $\cos(x^2)\cdot2x\,dx$ becomes $\cos u\,du$, and the integral becomes

$$\int\cos u\,du=\sin u+C=\sin(x^2)+C.$$

Whoever wrote $\int 2x\cos(x^2)$, without $dx$, has nowhere to put the $du$, and the $2x$ is left over for no reason. Leibniz’s notation did the work: the $dx$ knew how to turn into $du$.

Exercise: a Riemann sum by hand

Calculate the Riemann sum of $x^2$ on $[0,2]$ with $n=4$ strips, from the left and from the right.

$\Delta x=0.5$. From the left, the heights are $f(0)$, $f(0.5)$, $f(1)$, $f(1.5)$: $(0+0.25+1+2.25)\cdot0.5=1.75$. From the right, $f(0.5)$ to $f(2)$: $(0.25+1+2.25+4)\cdot0.5=3.75$. The true area, $\frac83\approx2.667$, lies between the two, as it should for an increasing function.

Exercise: let the dx do the work

Calculate $\int 3x^2(x^3+1)^4\,dx$.

With $u=x^3+1$, $du=3x^2\,dx$. The integral becomes $\int u^4\,du=\frac{u^5}{5}+C=\frac{(x^3+1)^5}{5}+C$. Checking: the derivative of $\frac{(x^3+1)^5}{5}$ is $(x^3+1)^4\cdot3x^2$, by the chain rule.

Chapter index · ← The D that factorises: the derivative as an operator · The fundamental theorem: when d and ∫ cancel →