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Areas under a hyperbola and the function ln x

In this article, Norman Wildberger explains a remarkable property of areas under the function y=1/x.
Graph showing the area under y=1/x from 1 to 2.5, and also showing the line y=lnx.
© UNSW Australia 2015
The function \(\normalsize{y=\frac{1}{x}}\) has some remarkable properties if we look at the areas underneath it. This question of how much area is under a particular graph is one of the two central problems of calculus. But sometimes we can derive surprising relations without calculus, as we do here.
In this step, we
  • look at a lovely symmetry about areas under the hyperbola \(\normalsize{y=\frac{1}{x}}\)
  • introduce the \(\ln(x)\) function in an elementary way without calculus
This is a more advanced step, but if you go carefully, then you can follow all the arguments, which are really quite spectacular in their consequences!

Rectangles under \(\normalsize{y=\frac{1}{x}}\)

First let’s note that any point \(\normalsize{[a,1/a]}\) on the graph of the hyperbola, say where \(\normalsize{a>0}\), determines a rectangle with the \(\normalsize{x}\) and \(\normalsize{y}\) axes. And the area of this rectangle is always exactly \(\normalsize{1}\), since the base and height of this rectangle are \(\normalsize{a}\) and \(\normalsize{1/a}\), which multiply to \(\normalsize{1}\).
Pictures of y=1/x with various rectangles of the same area
Q1 (E): What is the common area of such rectangles for the hyperbola \(\normalsize{y=\frac{2}{3x}}\)?
But other kinds of areas under this graph are also interesting, and exhibit an interesting property when we scale things.
pic of the graph y=1/x, showing area underneath it from 1 to 2, and from 2 to 4
Q2 (E): Explain why the two rectangular areas are equal.

A mixed dilation of the plane

Consider a transformation \(\normalsize{\phi}\) of the plane that sends a point \(\normalsize{P=[x,y]}\) to
\[\Large{\phi(P)=[2x,\frac{y}{2}]}.\]
This is a stretch in the horizontal direction by \(\normalsize{2}\), and at the same time a shrink by \(\normalsize{\frac{1}{2}}\) in the vertical direction. So \(\normalsize{\phi}\) does not change the areas of rectangles. And since general regions are made up (approximately) of rectangles, we can reason that although \(\normalsize{\phi}\) will change the shape of regions, it will not change their area.
But what do we mean by area of a region? Although we have a clear understanding of areas of rectangles, the definition of “area” for a curved region is much more problematic logically. The 17th century mathematicians already realised this. In practical applied mathematics we just assume that the term consistently extends to more complicated curved regions, but in pure mathematics trying to pin the term down embroils us in all kinds of subtle issues.
So let’s adopt a pragmatic attitude, not worry about the logical difficulties, and so assume that the word “area” really makes sense even for regions bounded by graphs like \(\normalsize{y=\frac{1}{x}}\).

The action of \(\normalsize{\phi}\) on the graph of \(\normalsize{y=\frac{1}{x}}\)

Let’s observe that points on the \(\normalsize{x}\)-axis get moved by \(\normalsize{\phi}\) to other points on the \(\normalsize{x}\)-axis, so that for example the interval \(\normalsize{1 \leq x \leq 2}\) will be sent to the interval \(\normalsize{2 \leq x \leq 4}\). The same holds for points on the \(\normalsize{y}\)-axis.
Now if \(\normalsize{A=[a,\frac{1}{a}]}\) is a point on the graph of \(\normalsize{y=\frac{1}{x}}\), then
\[\Large{\phi(A)=[2a,\frac{1}{2a}]}\]
is still a point on the graph. This means that the entire graph of \(\normalsize{y=\frac{1}{x}}\) is mapped to itself by \(\normalsize{\phi}\).
pic of y=1/x, with region under it from 1 to 2 shaded blue, and region from 2 to 4 shaded red
And now since our graph is actually preserved by \(\normalsize{\phi}\), the area under it from \(\normalsize{1}\) to \(\normalsize{2}\) must be the same as the area under it from \(\normalsize{2}\) to \(\normalsize{4}\). This is an elementary but remarkable observation!! We hope you have been able to follow the argument. Well done if you have!

An unbounded area!

If we apply \(\normalsize \phi\) again, we see the area under the graph from \(\normalsize{4}\) to \(\normalsize{8}\) is also exactly the same as the previous two areas! And similarly the area from \(\normalsize{8}\) to \(\normalsize{16}\) is the same, and from \(\normalsize{16}\) to \(\normalsize{32}\) is the same.
As we continue this, we conclude that the total area under the curve from \(\normalsize{1}\) onwards exceeds any known bound, or is unbounded. So we can state an important theoretical result: The area under the graph \(\normalsize{y=1/x}\) from \(\normalsize{x=1}\) to \(\normalsize{x=\infty}\) is unbounded.

Introducing \(\normalsize{\ln}\), as areas under \(\normalsize{y=\frac{1}{x}}\)

If you have followed this argument so far, we can go just a little bit further to obtain an even more remarkable relation. Suppose we consider the areas under \(\normalsize{y=\frac{1}{x}}\) over the three intervals \(\normalsize{[1,2]}\), \(\normalsize{[1,3]}\) and \(\normalsize{[1,6]}\). Let’s call these areas \(\normalsize{I(1,2)}\), \(\normalsize{I(1,3)}\) and \(\normalsize{I(1,6)}\).
The same argument we have already used shows that \(\normalsize{I(1,2)}\) is the same as \(\normalsize{I(3,6)}\), since this time we dilate and shrink by \(\normalsize{3}\) and \(\normalsize{\frac{1}{3}}\). So then
\[\Large{I(1,6)=I(1,3)+I(3,6)=I(1,3)+I(1,2)}.\]
The same argument applies for more general numbers:
\[\Large{I(1,a)+I(1,b)=I(1,ab)}.\]
In fact this is the same rule as satisfied by the \(\normalsize{\ln}\) function, often called the natural logarithm: namely
\[\Large{\ln\,a +\ln\,b =\ln\,ab}.\]
It turns out that this is more than just a coincidence, in fact
\[\Large{\ln\,a=I(1,a)}.\]
Areas under a hyperbola and the function ln x title
In fact in some treatments of calculus, the function \(\normalsize{\ln\;x}\) is defined precisely this way — as the area under \(\normalsize{y=\frac{1}{x}}\) from \(\normalsize{1}\) to \(\normalsize{x}\). So a large part of the calculus that deals with logarithm and exponential functions ultimately hinges on the remarkable hyperbola \(\normalsize{y=1/x}\) and its geometrical properties!
Q3 (E): What is the value of \(\normalsize{\ln\,1}\)?
Q4 (C): (for discussion) Is it possible to assign a certain value for \(\normalsize{\ln\;0}\)? What should this value be?
Q5 (C): (for discussion) how do you pronounce the function \(\ln\)?

Answers

A1. Since the typical point on the hyperbola \(\normalsize{y=\frac{2}{3x}}\) is \(\normalsize{[a,\frac{2}{3a}]}\) the area of rectangles is \(\normalsize{a\times\frac{2}{3a}=\frac{2}{3}}.\)
A2. The second rectangle has twice the base of the first, but half the height. So the areas must be equal!
A3. As the number \(\normalsize{\ln\,a}\) is defined as the area between the x axis and the graph of \(\normalsize{y=1/x}\) from \(\normalsize{1}\) to \(\normalsize{a}\), we deduce that
\[\Large{\ln\,1=0}.\]
A4. First of all for \(\normalsize{0<x<1}\) the value of \(\normalsize{\ln\,x}\) needs to be negative. The reason is a little bit subtle: actually we want to consider ‘signed areas’ and since we are going backwards from \(\normalsize{1}\) to \(\normalsize{x}\) the value is deemed to be negative. And what happens as \(\normalsize{0<x<1}\) gets closer and closer to \(\normalsize{0}\), but from the positive side? Well then the area under the graph gets to be a larger and larger negative number.
And why is that? It is because the curve \(\normalsize{y=1/x}\) is symmetric if we interchange \(\normalsize{x}\) and \(\normalsize{y}\), so the two areas, blue and red, shown here had better be the same – and our earlier argument showed that the blue area was unbounded. So the red one must also be unbounded!
pic of y=1/x with area from 1 to 10 coloured blue, and area from 0 to 1 above the line y=1 coloured red
A5. Norman says \(\ln\) is “lon”. Daniel says “natural logarithm”.
© UNSW Australia 2015
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Maths for Humans: Linear, Quadratic & Inverse Relations

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