Answer
$$
x=t^{2}-t, y=t^{4}, \quad 1 \leq t \leq 4
$$
the length of the given curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
& \approx 255.3756
\end{aligned}
$$
Work Step by Step
$$
x=t^{2}-t, y=t^{4}, \quad 1 \leq t \leq 4
$$
then
$$
d x / d t=2 t-1, \quad d y / d t=4 t^{3},
$$
so
$$
\begin{aligned}
(d x / d t)^{2}+(d y / d t)^{2}& =(2 t-1)^{2}+\left(4 t^{3}\right)^{2} \\
&=4 t^{2}-4 t+1+16 t^{6} .
\end{aligned}
$$
Thus the length of the given curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{1}^{4} \sqrt{16 t^{6}+4 t^{2}-4 t+1} d t \\
& \approx 255.3756
\end{aligned}
$$
