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Next: 8.6 Integrals and Area Up: 8. Integrable Functions Previous: 8.4 The Ruler Function   Index

8.5 Change of Scale

8.42   Definition (Stretch of a function.) Let $[a,b]$ be an interval in $\mbox{{\bf R}}$, let $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$, and let $f\colon
[a,b]\to\mbox{{\bf R}}$. We define a new function $f_r\colon [ra,rb]\to\mbox{{\bf R}}$ by

\begin{displaymath}f_r(t)=f({t\over r}) \mbox{ for all } t\in [ra,rb].\end{displaymath}

If $t\in [ra,rb]$, then $\displaystyle {{t\over r}\in [a,b]}$, so $\displaystyle {f({t\over r})}$ is defined.

The graph of $f_r$ is obtained by stretching the graph of $f$ by a factor of $r$ in the horizontal direction, and leaving it unstretched in the vertical direction. (If $r<1$ the stretch is actually a shrink.) I will call $f_r$ the stretch of $f$ by $r$.

8.43   Theorem (Change of scale for integrals.) Let $[a,b]$ be an interval in $\mbox{{\bf R}}$ and let $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$. Let $f\colon
[a,b]\to\mbox{{\bf R}}$ and let $f_r$ be the stretch of $f$ by $r$. If $f$ is integrable on $[a,b]$ then $f_r$ is integrable on $[ra,rb]$ and $\displaystyle { \int_{ra}^{rb}f_r=r\int_a^b f}$, i.e.,
\int_{ra}^{rb}f({x\over r})dx=r\int_a^b f(x)dx.
\end{displaymath} (8.44)

Proof: Suppose $f$ is integrable on $[a,b]$. Let $(\{P_n\},\{S_n\})$ be an arbitrary partition-sample sequence for $[ra,rb]$. If

\begin{displaymath}P_n=\{x_0,\cdots,x_m\} \mbox{ and } S_n=\{s_1,\cdots ,s_m\},\end{displaymath}


\begin{displaymath}{1\over r}P_n=\Big\{ {{x_0}\over r},\cdots ,{{x_m}\over r}\Bi...
...\over r}S_n=\Big\{ {{s_1}\over r},\cdots ,{{s_m}\over r}\Big\}.\end{displaymath}

Then $\Big(\Big\{ {1\over r}P_n\Big\},\Big\{ {1\over r}S_n\Big\}\Big)$ is a partition-sample sequence for $[a,b]$, so
${\{\sum(f,{1\over r}P_n,{1\over
r}S_n)\}\to \displaystyle {\int_a^b f}}$. Now

\sum(f_r,P_n,S_n) &=& \sum_{i=1}^m f_r(s_i) (x_i-x_{i-1}) \\
...i-1}}\over r}\Big)=r\sum \Big(f,{1\over r}P_n,{1\over r}S_n\Big)


\begin{displaymath}\lim\{\sum (f_r,P_n,S_n)\}=\lim\Big\{ r\sum(f,{1\over r}P_n,{1\over
r}S_n)\Big\}=r\int_a^b f.\end{displaymath}

This shows that $f_r$ is integrable on $[ra,rb]$, and $\displaystyle { \int_{ra}^{rb}f_r=r\int_a^b f}$. $\diamondsuit$

Remark: The notation $f_r$ is not a standard notation for the stretch of a function, and I will not use this notation in the future. I will usually use the change of scale theorem in the form of equation (8.44), or in the equivalent form

\int_A^B g(rx) dx = {1\over r} \int_{rA}^{rB}g(x)dx.
\end{displaymath} (8.45)

8.46   Exercise. A Explain why formula (8.45) is equivalent to formula (8.44).

8.47   Example. We define $\pi$ to be the area of the unit circle. Since the unit circle is carried to itself by reflections about the horizontal and vertical axes, we have

\begin{displaymath}\pi=4 \mbox{ (area (part of unit circle in the first quadrant)). } \end{displaymath}

Since points in the unit circle satisfy $x^2+y^2=1$ or $y^2=1-x^2$, we get

\begin{displaymath}\pi=4\int_0^1\sqrt{1-x^2}\; dx.\end{displaymath}

We will use this result to calculate the area of a circle of radius $a$. The points on the circle with radius $a$ and center 0 satisfy $x^2+y^2=a^2$, and by the same symmetry arguments we just gave

\mbox{ area(circle of radius } a) &=&4\int_0^a\sqrt{a^2-x^2}\;...
...nt_{a\cdot 0}^{a\cdot 1}\sqrt{1-\left({x\over a}\right)^2} \;dx.

By the change of scale theorem

\begin{displaymath}\mbox{ area(circle of radius } a)=4aa\int_0^1\sqrt{1-x^2} \;dx=a^2\pi.\end{displaymath}

The formulas

\begin{displaymath}\int_0^1\sqrt{1-x^2}\;dx ={\pi\over 4} \mbox{ and } \int_{-1}^1\sqrt{1-x^2}
\;dx={\pi\over 2}\end{displaymath}

or more generally

\begin{displaymath}\int_0^a\sqrt{a^2-x^2}\;dx={{\pi a^2}\over 4} \mbox{ and }
\int_{-a}^a\sqrt{a^2-x^2}={{\pi a^2}\over 2},\end{displaymath}

are worth remembering. Actually, these are cases of a formula you already know, since they say that the area of a circle of radius $a$ is $\pi a^2$.

8.48   Exercise. A Let $a,b$ be positive numbers and let $E_{ab}$ be the set of points inside the ellipse whose equation is

\begin{displaymath}{{x^2}\over {a^2}}+{{y^2}\over {b^2}}=1.\end{displaymath}

Calculate the area of $E_{ab}$.

8.49   Exercise. The figure shows the graph of a function $f$.
Let functions $g$, $h$, $k$, $l$, and $m$ be defined by
$\displaystyle {g(x) = f({x\over 3})}$.
$\displaystyle {h(x) = f(3x)}$.
$\displaystyle {k(x) = f({x+3 \over 3})}$.
$\displaystyle {l(x) = f(3x+3)}$.
$\displaystyle {m(x) = 3f({x \over 3})}$.
Sketch the graphs of $g$,$h$,$k$, $l$, and $m$ on different axes. Use the same scale for all of the graphs, and use the same scale on the $x$-axis and the $y$-axis,

8.50   Exercise. A The value of $\displaystyle {\int_0^1 {1\over 1+x^2}\;dx }$ is $.7854$(approximately). Use this fact to calculate approximate values for

\begin{displaymath}\int_0^a {1\over a^2 + x^2} dx \mbox{ and }\int_0^{{1\over a}}
{1\over 1 + a^2x^2} dx\end{displaymath}

where $a \in \mbox{${\mbox{{\bf R}}}^{+}$}$. Find numerical values for both of these integrals when $a = {1\over 4}$.

next up previous index
Next: 8.6 Integrals and Area Up: 8. Integrable Functions Previous: 8.4 The Ruler Function   Index
Ray Mayer 2007-09-07