On 29 October 1675 Leibniz was in Paris working on a problem about areas. Until then he had written omn., short for the Latin omnia, “all”, to mean the sum of all the lines of a figure, as the Italian Cavalieri did. Halfway down the page he wrote:

It will be useful to write ∫ instead of omn.

— Leibniz, manuscript of 29 October 1675

The symbol $\int$ is a long S, for summa, sum. At the time the long s was an ordinary letter in handwriting. It has never been replaced since.

Adding up slices

Leibniz’s idea was to cut a region into extremely thin vertical slices. Each slice is almost a rectangle, with height $y$ and width $dx$. Its area is $y\,dx$. The total area is the sum of all the slices:

$$\int y\,dx.$$

The symbol says it all at once: add up ($\int$) the heights ($y$) times the tiny widths ($dx$). The limits of the sum go below and above: $\displaystyle\int_a^b f(x)\,dx$ is the sum of the slices from $x=a$ to $x=b$.

Adding differences gives back the total

Remember the problem Huygens set Leibniz, told in the biography? There a sum of differences cancelled almost entirely:

$$(F_1-F_0)+(F_2-F_1)+(F_3-F_2)+\cdots+(F_n-F_{n-1})=F_n-F_0.$$

If you add up all the small changes of a quantity, only the total change is left: the final value minus the initial one. Leibniz realized the same happens with infinitely small differences. If $dF=f(x)\,dx$, then

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

In other words, $\int$ and $d$ undo each other, like adding and subtracting. This is the fundamental theorem of calculus, written as briefly as possible. Newton reached the same conclusion by the path of fluxions.

Example: the area under a parabola

For the area under $y=x^2$ between $0$ and $3$, we look for $F$ with $dF=x^2\,dx$. Since $d\!\left(\frac{x^3}{3}\right)=x^2\,dx$:

$$\int_0^3x^2\,dx=\frac{3^3}{3}-\frac{0^3}{3}=9.$$

Example: distance travelled

A car accelerates with velocity $v(t)=3t^2$ metres per second. In each tiny piece of time $dt$ it travels $v\,dt$. Between $t=0$ and $t=2$ seconds, the total distance is

$$\int_0^2 3t^2\,dt=\Big[t^3\Big]_0^2=8\ \text{metres}.$$

The units help to check: velocity (m/s) times time (s) gives metres. Leibniz’s notation carries the units along, which is one reason it is so widely used in physics and engineering.

A word of caution: if the velocity changes sign, that is, if the car reverses, the integral gives the displacement, the final position minus the initial one. For the total distance travelled, we integrate the absolute value of the velocity.

The triangle Pascal did not see

In 1673, newly arrived in Paris, Leibniz read a text by Blaise Pascal on the sines of a quarter circle, published in the Letters of A. Dettonville. At a point on the circle, Pascal drew a tiny triangle with one horizontal side, one vertical side and the third side along the curve.

Leibniz realized that the triangle worked for any curve, not just the circle. Its sides are $dx$, $dy$ and a little piece $ds$ of the curve itself. Since the triangle is right-angled, Pythagoras gives

$$ds^2=dx^2+dy^2.$$

He called it the characteristic triangle and later said that this was where he saw a light that Pascal himself had missed. With this triangle Leibniz found tangents (the slope is $dy/dx$) and also lengths of curves: just add up the $ds$.

$$L=\int ds=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$

For example, for $y=\frac23x^{3/2}$ between $x=0$ and $x=3$, we have $\dfrac{dy}{dx}=\sqrt x$, and the length is

$$L=\int_0^3\sqrt{1+x}\,dx=\Big[\tfrac23(1+x)^{3/2}\Big]_0^3=\tfrac23(8-1)=\tfrac{14}{3}.$$

The notation does the work: Leibniz’s $ds$ is the hypotenuse, and the integral is the sum of all of them.

The first publication

Leibniz published the symbol $\int$ in 1686, in an Acta Eruditorum paper called On a hidden geometry. There he showed how areas and tangents, two problems that seemed different, were two sides of the same coin.

Exercise: compute the integral of 2x + 1 between 0 and 3

An antiderivative is $x^2+x$, because $d(x^2+x)=(2x+1)\,dx$. The integral is $(9+3)-(0+0)=12$. Since the function is positive on the interval, this number is also the area under the graph.

Exercise: a tap delivers 4t litres per minute. How much does it fill between t = 0 and t = 5 minutes?

$\displaystyle\int_0^5 4t\,dt=\Big[2t^2\Big]_0^5=50$ litres.

Sources: Leibniz, manuscripts and publications, MacTutor.

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