Stretching an integral

Stretching an integral

Excerpt: How can we integrate $\frac{1}{x}$? Students are asked what happens to areas under the graph of $y=\frac{1}{x}$ when it is stretched, leading to a formula for the integral without requiring calculus tools.

Warm-up

The graph of the function $y=\dfrac{1}{x^2}$ is stretched as shown in this diagram.

If the graph is stretched by a factor of $b$ in the $x$-direction (as shown), what is the equation of the resulting graph?

And if, instead, the original graph is stretched by a factor of $c$ in the $y$-direction, what is the equation of the resulting graph?

Warm-up solution

The graph of the function $y=\dfrac{1}{x^2}$ is stretched as shown in this diagram.

If the graph is stretched by a factor of $b$ in the $x$-direction (as shown), what is the equation of the resulting graph?

One way to think about this problem is to use the language of functions.

The original graph has equation $y=f(x)$, where $f(x)=\dfrac{1}{x^2}$.

When we stretch the graph by a factor of $b$ in the $x$-direction, a point with coordinates $(x,f(x))$ is stretched to the point with coordinates $(bx,f(x))$, as shown in the following diagram. If we write $X=bx$ for the new $x$-coordinate, then the new point has coordinates $(X,Y)=(X,f(X/b))$, so the new graph has equation $Y=f(X/b)$, or $y=f(x/b)$ using lower-case letters.

So the equation of the stretched graph is $y=f(x/b)=\dfrac{1}{(x/b)^2}$. Multiplying the numerator and denominator by $b^2$ gives the simpler form $y=\dfrac{b^2}{x^2}$.

And if, instead, the original graph is stretched by a factor of $c$ in the $y$-direction, what is the equation of the resulting graph?

This time, the point $(x,f(x))$ is stretched to the point with coordinates $(x,cf(x))$, as shown in the following diagram. Writing the coordinates of the new point again as $(X,Y)=(x,cf(x))$, we find the new graph has equation $Y=cf(X)$, as $X=x$, or $y=cf(x)$ using lower-case letters.

Therefore the equation of the stretched graph is $y=\dfrac{c}{x^2}$.

Problem

Take a look at these diagrams. (There is a printable version of them available.)

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.

Interactive applet

In the following applet, you can:

• choose the value of $a$

• stretch the graph (and the shaded region with it) in the $x$-direction by a scale factor of $b$

• stretch the graph (and the shaded region with it) in the $y$-direction by a scale factor of $c$

The original area $I(a)$ is shown, and the area of the stretched shaded region is also shown.

The original (unstretched) graph of $y=\dfrac{1}{x}$ is shown as a dashed red line.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/xa8hvBNg/width/584/height/398/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

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

Recall from above that $I(x)$ is the function that gives the area of the shaded region between $1$ and $x$ under the graph of $y = \dfrac1x$, so $I(x)=\displaystyle{\int_1^x \frac{1}{t}\,dt}$. (To avoid confusion, we have used $t$ when writing this integral instead of $x$, since $x$ appears in the limits.)

• 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?

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

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? > [interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/RTkHVBpq/width/577/height/396/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page. Download GeoGebra file

• What happens if one or both of $a$ and $b$ are less than $1$?

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.

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

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?