13.3 Integrals of Intervals
By computing the integral of an interval, we can measure the interval's length.
Definition.
Suppose we have a ruler and an interval
\[
\begin{align}
\bar{\diff} z &: \Ruler_{x}
&
i &: \Interval_{x}
\end{align}
\]
where \( \bar{\diff} z \) is the differential of a function \( z \depends x \). The integral of \( i \)
\[
\begin{align}
\Bigl\langle{\bar{\diff} z}\mathrel{\textstyle{\int}}{i}\Bigr\rangle &: \Number
\\
\Bigl\langle{\bar{\diff} z}\mathrel{\textstyle{\int}}{i}\Bigr\rangle &= \Bigl\langle{z}\mathrel{\textstyle{\int}}{\boundary i}\Bigr\rangle
\end{align}
\]
is computed by trading the differential, \( \bar{\diff} \), for a boundary, \( \boundary \).
The standard ruler
\[
\begin{align}
\bar{\standard}_{x} &: \Ruler_{x}
\end{align}
\]
is also known as the length ruler for the \( x \)-coordinate line.
Example.
Let's use the length ruler, \( \bar{\standard}_{x} \), to measure the interval \( [1, 5]_{x} \). We compute.
\[
\begin{align}
\biggl\langle{\bar{\standard}_{x}}\mathrel{\displaystyle{\int}}{[1, 5]_{x}}\biggr\rangle &= \biggl\langle{\bar{\diff} x}\mathrel{\displaystyle{\int}}{[1, 5]_{x}}\biggr\rangle
\\
&= \biggl\langle{x}\mathrel{\displaystyle{\int}}{\boundary [1, 5]_{x}}\biggr\rangle
\\
&= \biggl\langle{x}\mathrel{\displaystyle{\int}}{[5]_{x} - [1]_{x}}\biggr\rangle
\\
&= \biggl\langle{x}\mathrel{\displaystyle{\int}}{[5]_{x}}\biggr\rangle - \biggl\langle{x}\mathrel{\displaystyle{\int}}{[1]_{x}}\biggr\rangle
\\
&= \biggl\langle{5}\mathrel{\displaystyle{\int}}{\sliced}\biggr\rangle - \biggl\langle{1}\mathrel{\displaystyle{\int}}{\sliced}\biggr\rangle
\\
&= 4
\end{align}
\]
The length integral just had us subtract one from five. That's fair enough!
Integrals become more interesting when our rulers include dependence.
Example.
Consider the function
\[
\begin{align}
z &\depends x
\\
z &= x^2.
\end{align}
\]
Let's compute the integral
\[
\begin{align}
&\biggl\langle{\bar{\diff} z}\mathrel{\displaystyle{\int}}{[1, 2]_{x}}\biggr\rangle
\end{align}
\]
that measures the interval \( [1, 2]_{x} \) with the ruler \( \bar{\diff} z \). We find
\[
\begin{align}
\biggl\langle{\bar{\diff} z}\mathrel{\displaystyle{\int}}{[1, 2]_{x}}\biggr\rangle &= \biggl\langle{z}\mathrel{\displaystyle{\int}}{\boundary [1, 2]_{x}}\biggr\rangle
\\
&= \biggl\langle{z}\mathrel{\displaystyle{\int}}{[2]_{x} - [1]_{x}}\biggr\rangle
\\
&= \biggl\langle{z}\mathrel{\displaystyle{\int}}{[2]_{x}}\biggr\rangle - \biggl\langle{z}\mathrel{\displaystyle{\int}}{[1]_{x}}\biggr\rangle
\\
&= \biggl\langle{x^2}\mathrel{\displaystyle{\int}}{[2]_{x}}\biggr\rangle - \biggl\langle{x^2}\mathrel{\displaystyle{\int}}{[1]_{x}}\biggr\rangle
\\
&= \biggl\langle{4}\mathrel{\displaystyle{\int}}{\sliced}\biggr\rangle - \biggl\langle{1}\mathrel{\displaystyle{\int}}{\sliced}\biggr\rangle
\\
&= 3.
\end{align}
\]
This is not the measurement we would find using the length ruler, \( \bar{\standard}_{x} \).
What role did rulers play in previous example's integral? On the \( x \)-axis, the ruler
\[
\begin{align}
\bar{\diff} z &= 2 x \mult \bar{\standard}_{x}
\end{align}
\]
is non-standard. We would like to measure the interval \( [1, 2]_{x} \) with this ruler. But to make the measurement, we should switch to \( z \)-coordinates! On the \( z \)-axis
\[
\begin{align}
\bar{\diff} z &= \bar{\standard}_{z}
\end{align}
\]
is the standard ruler.
Measuring the corresponding interval \( [1, 4]_{z} \) just computes the change in height.
\[
\begin{align}
\change_{z} &= 4 - 1 = 3
\end{align}
\]
What would you expect to find if we measured the interval \( [-1, 2]_{x} \) instead? Or the interval \( [-1, 1]_{x} \)?