Measuring the area of curved figures is one of the oldest problems in mathematics. Around 250 BC Archimedes showed that the area of a parabolic segment is $\frac43$ of the area of a triangle inscribed in it. It was a stroke of genius, but made to measure: every new curve needed a new idea.

By the 17th century, Fermat, Cavalieri, Wallis and others could already compute the area under $y=x^n$ for many values of $n$. What was missing was a general rule. And what was missing was the realization that Newton and Leibniz each reached in their own way: computing areas and computing rates of change are inverse operations.

Area as something that grows

Think of the area under the curve $y=f(x)$ from $0$ to a point $x$. Call it $A(x)$. If you push the right edge from $x$ to $x+h$, the area gains a thin strip.

That strip is almost a rectangle of height $f(x)$ and width $h$. So

$$A(x+h)-A(x)\approx f(x)\cdot h\qquad\Longrightarrow\qquad\frac{A(x+h)-A(x)}{h}\approx f(x).$$

The thinner the strip, the better the approximation. In the limit:

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

In Newton’s language: the fluxion of the area is the height of the curve. The area grows exactly at the rate given by the height. To find the area, we just need a function whose fluxion is $f$, that is, to go the inverse way.

To be rigorous, we trap the strip between two rectangles: one with the smallest height of $f$ on the interval and one with the largest. If $f$ is continuous, both heights approach $f(x)$ as $h\to0$, and the strip is squeezed between them.

The rule of De analysi

In 1669 Newton wrote a short treatise called De analysi. Its first rule said, in today’s notation, that the area under $y=ax^{m/n}$, from $0$ to $x$, is

$$\frac{an}{m+n}\,x^{(m+n)/n}.$$

It looks complicated, but it is the rule you know: the area under $x^k$ is $\dfrac{x^{k+1}}{k+1}$. For example, under $y=x^2$ between $0$ and $2$:

$$\int_0^2 x^2\,dx=\left[\frac{x^3}{3}\right]_0^2=\frac{8}{3}.$$

Archimedes would have needed a whole proof to reach that number. With the rule, it takes one line.

The logarithm as an area

Combining the rule with the generalized binomial, Newton computed areas nobody knew how to compute. The hyperbola $y=\dfrac{1}{1+x}$ becomes a sum of powers:

$$\frac{1}{1+x}=1-x+x^2-x^3+\cdots$$

Now we just compute the area under each power, term by term:

$$\int_0^x\frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$

That area is the natural logarithm $\ln(1+x)$. With $x=0.1$, four terms give $0.0953083$, and the correct value is $0.0953102$.

What it is for

Integrating velocity gives the distance travelled. Integrating a density gives a mass. Integrating a flow rate gives the volume that filled the tank. In all these cases we know how something changes at each instant and want to know how much it changed in total. Calculus links the two questions: how does it change? and how much has it changed?

Exercise: compute the area under y = 2x between 0 and 3

An antiderivative of $2x$ is $x^2$. The area is $3^2-0^2=9$. Check it with geometry: it is a triangle with base 3 and height 6, whose area is $\frac{3\cdot6}{2}=9$.

Exercise: use Newton's rule for the area under y = √x between 0 and 4

$\sqrt x=x^{1/2}$, so the area is $\dfrac{x^{3/2}}{3/2}=\dfrac23x^{3/2}$. Between $0$ and $4$: $\dfrac23\cdot8=\dfrac{16}{3}$.

Sources: mathematical manuscripts at the Newton Project · MacTutor biography.

All Isaac Newton topics · ← Fluxions: how to measure an instantaneous change · Newton’s method: approximating roots with tangents →

Image credits.