Before calculus, tangents and areas were two unrelated problems. Finding the tangent to a curve was one art; finding the area beneath it, another. The great discovery of the seventeenth century, by Isaac Barrow, Newton and Leibniz, was that one is the inverse of the other. This is the fundamental theorem of calculus, and Leibniz’s notation writes it in a way that makes it look almost obvious.

Leibniz: like sums and differences

Leibniz thought of $\int$ as a sum and of $d$ as a difference. And adding up differences undoes the difference, as on a staircase: the sum of the steps is the total height. He wrote that $\int$ and $d$ are inverse to each other, “like powers and roots in ordinary calculation”. In symbols:

$$d\!\int y\,dx=y\,dx,\qquad\int dy=y.$$

The $d$ and the $\int$ “cancel”, like $+3$ and $-3$. It is an active notation in the strongest sense: the most important theorem of calculus appears as a simplification rule.

The two halves, in modern notation

Today the theorem is written in two parts.

First part: differentiate the area. If $A(x)=\displaystyle\int_a^x f(t)\,dt$ is the area accumulated from $a$ to $x$, then

$$\frac{d}{dx}\int_a^x f(t)\,dt=f(x).$$

The rate at which the area grows is the height of the curve at the point where you are. Notice that the letter inside the integral is $t$, not $x$: the $x$ is already taken as the upper limit. Writing $\int_a^x f(x)\,dx$ would be the variable capture mistake from chapter 3.

Second part: integrate a derivative. If $F'=f$, then

$$\int_a^b f(x)\,dx=F(b)-F(a)=\Big[F(x)\Big]_a^b.$$

The bracket with limits is a transitional notation: it holds $F$ while the numbers are still to be substituted. With it, the area that in the previous text required infinitely many rectangles comes out in one line:

$$\int_0^2x^2\,dx=\left[\frac{x^3}{3}\right]_0^2=\frac83-0=\frac83.$$

What the notation hides

The cancellation of $d$ with $\int$ is not perfect. In the indefinite integral,

$$\int\frac{dF}{dx}\,dx=F(x)+C,$$

a constant $C$ appears that the cancellation does not announce. Differentiating erases constants; integrating does not know which constant to give back. Forgetting the $+C$ is the most common mistake in calculus courses, and its origin is semiotic: the notation suggests a perfect inverse where there is only an inverse up to a constant.

Exercise: differentiate an area with a variable limit

Calculate $\dfrac{d}{dx}\displaystyle\int_0^{x^2}\cos t\,dt$.

Let $A(u)=\int_0^u\cos t\,dt$, with $A'(u)=\cos u$. The expression is $A(x^2)$, and by the chain rule its derivative is $A'(x^2)\cdot2x=2x\cos(x^2)$. Checking with the second part: $\int_0^{x^2}\cos t\,dt=\sin(x^2)$, whose derivative is $2x\cos(x^2)$.

Exercise: the +C in action

Two people integrate $2\sin x\cos x$. One gets $\sin^2x$; the other, $-\cos^2x$. Who is wrong?

Neither. $\sin^2x-(-\cos^2x)=\sin^2x+\cos^2x=1$: the two answers differ by a constant, and both are correct antiderivatives. A third person might get $-\frac12\cos2x$, which also differs from the others by a constant. That is what the $+C$ is for.

Chapter index · ← The integral read sign by sign · Wallis’s ∞: a sign that is not a number →