Take a look at these diagrams.

Four diagrams which are described in the text below

Diagram 1 shows the graph of \(y=\dfrac1x\). \(I(a)\) is the shaded region under the graph between \(x=1\) and \(x=a\), where \(a>1\), so \(I(a)=\displaystyle{\int_1^a \dfrac{1}{x}\,dx}\).

Step 1: The shaded region in Diagram 4 has the same area as the shaded region in Diagram 1 (\(I(a)\)) after undergoing two stretches (either via Diagram 2 or via Diagram 3). Fill in the following information on each diagram where it is missing:

  • the scale factor that was used in the stretch;

  • the \(x\)-coordinate at each end of the shaded region;

  • the area of the shaded region in terms of \(I(a)\), and

  • the equation of the graph.

Using the ideas from the Warm-up problem, we can complete the diagrams as follows:

The image as above with the scale factors and so on filled in

Step 2: Investigating the function \(I(x)\).

  • Looking now at Diagram 4, can you identify the area \(I(b)\) on the diagram?

  • Are there any other \(I(x)\) values which you can easily identify on the diagram?

  • How are \(I(a)\) and \(I(b)\) related?

Recall that \(I(b)\) is the area under the graph between \(x=1\) and \(x=b\), which we can see fits on the diagram very easily, giving the following:

Areas I of a and I of b pointed out on the plots.

The other \(I(x)\) value which is now easy to identify on the diagram is the whole shaded area from \(x=1\) to \(x=ab\), which is \(I(ab)\).

Putting these together, we see that \[I(ab) = I(a)+I(b).\]

In terms of integrals, this says that \[\int_1^{ab} \frac{1}{x}\,dx=\int_1^a \frac{1}{x}\,dx+\int_1^b \frac{1}{x}\,dx,\] which is far from obvious from the above sketch.

What do these answers suggest about the function \(I(x)\)? Have you seen anything with these properties before?

This suggests that \(I(x)\) may be a logarithm, because logarithms also have the property that \[\log ab=\log a + \log b.\]

Some further questions

  • Using this applet or some graphing software, investigate values of \(I(x)\). Which value of \(a\) gives \(I(a)=1\)? What does this tell you?

Experiment shows that \(I(a)=1\) when \(a\approx 2.72\). Therefore if \(I(x)\) really is a logarithm, the base of the logarithm is approximately \(2.72\).

We can actually use this result to define logarithms: we define the natural logarithm of \(x\), usually written \(\ln x\), to be \(I(x)\), so \[\ln x=\int_1^x \frac{1}{t}\,dt.\]

The base of these logarithms is called “\(e\)”, so we have discovered that \(e\approx 2.72\). Therefore, the natural logarithm can also be written as \(\log_e x\).

If you have already met the natural logarithm function, how can you check that the definition given here and your existing definition agree?

  • What happens if \(a\) and/or \(b\) is less than \(1\)?

The relationship we have discovered, \(I(ab)=I(a)+I(b)\) does still work, but we have to require that if \(a<1\), then \(I(a)\) is considered to be negative. This fits with the integral definition of \(I(a)\), that is \(I(a)=\displaystyle{\int_1^a \frac{1}{x}\,dx}\). Then if \(a<1\), the limits of the integral are in the “wrong order”, so we take the value of the integral to be \(\displaystyle{-\int_a^1 \frac{1}{x}\,dx}\) (with the limits in the “right order”), and this is negative. This is because we always require \[\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx.\]

Why might we require integrals to behave in this way?