# Areas under a hyperbola and the function ln x

- 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

## 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 \(\normalsize{1}\).Q1(E): What is the common area of such rectangles for the hyperbola \(\normalsize{y=\frac{2}{3x}}\)?

Q2(E): Explain why the two rectangular areas are equal.

## A mixed dilation of the plane

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

## An unbounded area!

*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 \(\normalsize{y=\frac{1}{x}}\)

*natural logarithm*: namely

*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!\[\Large{\ln\,1=0}.\]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 thatA4.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!A5.Norman says \(\ln\) is “lon”. Daniel says “natural logarithm”.

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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