Let's compute the path integral
\[
\begin{align}
&\Bigl\langle{r}\mathrel{\overset{p}{\textstyle{\int}}}{i}\Bigr\rangle
\end{align}
\]
with ruler, path, and interval given as follows.
\[
\begin{align}
r &: \Ruler_{(x, y)}
&
p &: \Path_{(x, y)}
&
i &: \Interval_{t}
\\
r &= y \mult \bar{\standard}_{x} + (2 y - x) \mult \bar{\standard}_{y}
&
&\left\{\begin{aligned} x_p &= t^2 \\ y_p &= t \end{aligned}\right.
&
i &= [0, 3]_{t}
\end{align}
\]
You can check that the ruler \( r \) is wild. So there is no hope of finding a function \( z \depends (x, y) \) that satisfies
\[
\begin{align}
\bar{\diff} z &= y \mult \bar{\standard}_{x} + (2 y - x) \mult \bar{\standard}_{y}.
\end{align}
\]
Let's instead use our path \( p \) to rewrite \( r \) as a ruler that measures time.
\[
\begin{align}
r &= y \mult \bar{\diff} x + (2 y - x) \mult \bar{\diff} y
\\
&= t \mult \bar{\diff} \bigl(t^2\bigr) + \bigl(2 t - t^2\bigr) \mult \bar{\diff} t
\\
&= t \mult 2 t \mult \bar{\diff} t + 2 t \mult \bar{\diff} t - t^2 \mult \bar{\diff} t
\\
&= \bigl(2 t^2 + 2 t - t^2\bigr) \mult \bar{\standard}_{t}
\\
&= \bigl(t^2 + 2 t\bigr) \mult \bar{\standard}_{t}
\end{align}
\]
To solve our integral, we should look for a function \( z \depends t \) that satisfies
\[
\begin{align}
\bar{\diff} z &= \bigl(t^2 + 2 t\bigr) \mult \bar{\standard}_{t}.
\end{align}
\]
By anti-differentiating, we find
\[
\begin{align}
\bigl(t^2 + 2 t\bigr) \mult \bar{\standard}_{t} &= \bigl(t^2 + 2 t\bigr) \mult \bar{\diff} t
\\
&= t^2 \mult \bar{\diff} t + 2 t \mult \bar{\diff} t
\\
&= \frac{1}{3} \mult 3 t^2 \mult \bar{\diff} t + 2 t \mult \bar{\diff} t
\\
&= \frac{1}{3} \mult \bar{\diff} \bigl(t^3\bigr) + \bar{\diff} \bigl(t^2\bigr)
\\
&= \bar{\diff} \biggl(\frac{1}{3} \mult t^3\biggr) + \bar{\diff} \bigl(t^2\bigr)
\\
&= \bar{\diff} \biggl(\frac{1}{3} \mult t^3 + t^2\biggr).
\end{align}
\]
We can calculate our path integral as a time integral.
\[
\begin{align}
\Bigl\langle{r}\mathrel{\overset{p}{\textstyle{\int}}}{i}\Bigr\rangle &= \biggl\langle{y \mult \bar{\standard}_{x} + (2 y - x) \mult \bar{\standard}_{y}}\mathrel{\overset{p}{\displaystyle{\int}}}{[0, 3]_{t}}\biggr\rangle
\\
&= \biggl\langle{\bigl(t^2 + 2 t\bigr) \mult \bar{\standard}_{t}}\mathrel{\displaystyle{\int}}{[0, 3]_{t}}\biggr\rangle
\\
&= \biggl\langle{\bar{\diff} \biggl(\frac{t^3}{3} + t^2\biggr)}\mathrel{\displaystyle{\int}}{[0, 3]_{t}}\biggr\rangle
\\
&= \biggl\langle{\frac{t^3}{3} + t^2}\mathrel{\displaystyle{\int}}{\boundary [0, 3]_{t}}\biggr\rangle
\\
&= \biggl\langle{\frac{t^3}{3} + t^2}\mathrel{\displaystyle{\int}}{[3]_{t}}\biggr\rangle - \biggl\langle{\frac{t^3}{3} + t^2}\mathrel{\displaystyle{\int}}{[0]_{t}}\biggr\rangle
\\
&= \biggl(\frac{27}{3} + 9\biggr) - (0 + 0)
\\
&= 18
\end{align}
\]
We've found our path integral!