Answer
The length of the curve
$$
y=4\ (x-1)^{\frac{3}{2}}, \quad\quad 1 \leq x \leq 4.
$$
is equal to $ \frac{1}{54} \left(109\sqrt{109}-1\right). $
Work Step by Step
$$
y=4\ (x-1)^{\frac{3}{2}}, \quad\quad 1 \leq x \leq 4.
$$
We have
$$
y=4\ (x-1)^{\frac{3}{2}} \Rightarrow \frac{d y}{d x}=6( x-1)^{\frac{1}{2}}
$$
$$
\Rightarrow 1+(d y / d x)^{2}=1+\left(6( x-1)^{\frac{1}{2}}\right)^{2}= 36x-35
$$
Thus, the arc length is
$$
\begin{aligned}
L& =\int_{1}^{4} \left( \sqrt {36x-35 }\right) d x \\
& =\left[\frac{1}{54} (36x-35)^{\frac{3}{2}}\right]_{1}^{4} \\
&=\left(\frac{1}{54} 109\sqrt{109}-1\right) \\
\end{aligned}
$$
Hence the length of the curve
$$
y=4\ (x-1)^{\frac{3}{2}}, \quad\quad 1 \leq x \leq 4.
$$
is equal to $ \frac{1}{54} \left(109\sqrt{109}-1\right). $
