# Stretching an integral Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Solution

Take a look at these diagrams.

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:

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:

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?