How Did Leibniz Derive the Formula to Solve the Area Problem in Calculus?

 by Belinda Chau

In calculus, one of the three central problems is the area problem which is to determine the area under the curve. The method for solving this is something a lot of people are already familiar with. However, when I was reading the book ‘Infinite Powers’ by Steven Strogatz, I was completely fascinated with how Leibniz, the mathematician who discovered calculus alongside Newton, derived the formula we use today to solve this type of problems.

This all began with a puzzle that Christiaan Huygens introduced Leibniz to, asking what was the sum to the following infinite number of terms:

112+123+134+ ...+ 1n(n+1)+ ...= ?

A way to work out what we can do is to simplify this question by limiting the number of terms to, for example, 99 terms, so we get:

 S=112+123+134+...+1n(n+1)+...+199100

The question has its clever trick to solve, and it was spotted by Leibniz: each term can be written as the difference of two consecutive unit fractions. For example:

First term =112 =2-112=11-12

Ninety-ninth term = 199100 =100-9999100=199-1100 

nth term =1n(n+1)=n+1-nn(n+1)=1n-1n+1

Having been able to notice this, we can rewrite the sum of all 99 terms as:

S= 11-12+12-13+13+...-199+199-1100=11-1100= 99100

Since every other term apart from 11 and 1100 can be cancelled out, S can be calculated by doing the first term minus the last term. This is how the trick conveniently converts the sum of this long list of terms in a single fraction; it is this property that makes S a telescoping sum. Using the same technique, we can solve our original question: if n tends to infinity, the term 1n+1 will tend to zero and the sum will tend to one, so the resulting limiting value of the sum will be one.

Leibniz wondered whether it was possible to incorporate telescoping in some way in the approximation of the area under a curve given its ability to simplify complex problems. The first step is to imagine slicing the finite region bounded by the curve and endpoints a and b (let’s call the area R) into eight vertical, rectangular strips. All strips have the same width of x and varying heights of y1, y2...y8 (see figure 1).

Figure 1. Area under the graph sliced into eight vertical, rectangular strips

At this stage, we are trying to figure out the pattern by working with a simplified version so that we can have a similar approach to more complex problems while achieving accuracy. To approximate the total area of R, we form the expression y1x+y2x+...+y8x to show it is the sum of the area of all strips with each term calculated using the formula base times height. Since the sum consists of a long list of terms, Leibniz thought we could use telescoping to express each term as the difference of two numbers with the numbers denoted by A0,A1...A8 which work as follow:

y1x=A1-A0

y2x=A2-A1

…

y8x=A8-A7

Adding up all these terms will result in a telescoping sum for the area of R where its expression simplifies to A8-A0, and the significance of these numbers will be revealed later.

Now, to accurately calculate the area of R, we need to imagine slicing R into infinitely many infinitesimally thin vertical, rectangular strips which neatly cover the entire area of R. In the limit of infinitesimally thin strips, their width is denoted with the differential dx. As the strips have varying height that is determined by the x value, height is given as the function y(x) to demonstrate this relationship. Therefore, our new expression for the area of each strip becomes y(x)dx, which allows us to form our expression for area of R to be the integral y(x)dx, meaning the area of R is the sum of the area of all strips. 

Before, we saw in our simplified version how the total area of the strips can be expressed as A8-A0 by giving the sum this telescoping property. Applying it to our area R consisting of infinitely many infinitesimally thin strips bounded by endpoints a and b with a<b, we get the area of R equals to A(b)-A(a). The endpoints a and b are our limits of integration, so we can derive the formula for calculating area of R to be:

 aby(x)dx=A(b)-A(a)

To determine the function A(x), from our previous equation y1x=A0-A1 we can derive the equation y(x)dx=dA for the infinitesimally thin strips where the differential dA denotes the infinitesimal change in area. If we then divide both sides of the equation by dx, we will get dAdx=y(x). Therefore, in the limit of infinitesimally thin strips, y(x) is the derivative of the function A(x), which can be found by integrating y(x) with respect to x. 

This is Leibniz’s method for deriving the formula to solve the area problem in calculus. For me, it is the creativity to come up with such a clever method of using telescoping to tackle a problem mathematicians had been trying to solve for centuries that amazes me. I believe learning about mathematical proofs can really take our understanding of maths to another level; it has the power to help us be clear and confident about the reasoning behind everything we do in our calculations and think logically through challenging problems. This is why understanding mathematical proofs is one of things I most enjoy about maths.

Reference

Strogatz. S. (2020) Infinite Powers: The Story of Calculus - The Language of the Universe. London: Atlantic Books