Answer
Thus the exact length of the curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=2(2 \sqrt{2}-1).
\end{aligned}
$$
Work Step by Step
$$
x=1+3 t^{2}, \quad y=4+2 t^{3}, \quad 0 \leq t \leq 1
$$
then
$$
d x / d t=6 t , \quad d y / d t=6 t^{2}
$$
so
$$
(d x / d t)^{2}+(d y / d t)^{2} =36 t^{2}+36 t^{4}
$$
Thus the exact length of the curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{0}^{1} \sqrt{36 t^{2}+36 t^{4}} d t \\
&=\int_{0}^{1} 6 t \sqrt{1+t^{2}} d t \\
& \quad \quad \quad \left[\text {Let } u=1+t^{2}, \text {then }d u=2 t d t \right]\\
&=6 \int_{1}^{2} \sqrt{u}\left(\frac{1}{2} d u\right) \\
&=3\left[\frac{2}{3} \vec{u}^{3 / 2}\right]_{1}^{2} \\
& =2\left(2^{3 / 2}-1\right) \\
&=2(2 \sqrt{2}-1).
\end{aligned}
$$
