Answer
The electric potential at $P$ is $V\left( P \right) = 16856.3$ volts.
Work Step by Step
We have the charge density $\delta \left( {x,y} \right) = {x^3}y$. The curve can be parametrized by
${\bf{r}}\left( t \right) = \left( {t,{t^{ - 1}}} \right)$, ${\ \ \ }$ for $\frac{1}{2} \le t \le 2$
Thus,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1, - {t^{ - 2}}} \right)\cdot\left( {1, - {t^{ - 2}}} \right)} = \sqrt {1 + {t^{ - 4}}} $
$\delta \left( {{\bf{r}}\left( t \right)} \right) = {t^2}$
The distance ${d_P}$ from $P = \left( {0,0} \right)$ to a point $\left( {x,y} \right)$ on the curve is
${d_P} = \sqrt {{x^2} + {y^2}} = \sqrt {{t^2} + {t^{ - 2}}} = t\sqrt {1 + {t^{ - 4}}} $
Using Eq. (6), the electric potential at $P$ is given by
$V\left( P \right) = k\mathop \smallint \limits_C^{} \frac{{\delta \left( {x,y} \right){\rm{d}}s}}{{{d_P}}} = k\mathop \smallint \limits_{1/2}^2 \frac{{\delta \left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }}{\rm{d}}s$
Using Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$,
where in this case $f\left( {x,y} \right) = \delta \left( {x,y} \right) = {x^3}y$, we obtain
$V\left( P \right) = k\mathop \smallint \limits_{1/2}^2 \frac{{{t^2}\sqrt {1 + {t^{ - 4}}} }}{{t\sqrt {1 + {t^{ - 4}}} }}{\rm{d}}t$
$V\left( P \right) = k\mathop \smallint \limits_{1/2}^2 t{\rm{d}}t = \frac{k}{2}\left( {{t^2}|_{1/2}^2} \right) = \frac{k}{2}\left( {4 - \frac{1}{4}} \right) = \frac{{15}}{8}k$
With the charge density in units of ${10^{ - 6}}$ and $k = 8.99 \times {10^9}$, we obtain
$V\left( P \right) = \frac{{15}}{8} \times 8.99 \times {10^9} \times {10^{ - 6}} = 16856.3$
So, the electric potential at $P$ is $V\left( P \right) = 16856.3$ volts.
