Answer
The length for the given curve is given by
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{1+\left[y^{\prime}(t)\right]^{2}} d t \\
&=\int_{1}^{4} \sqrt{x^{3}} d x \\
&=12.4
\end{aligned}
$$
Work Step by Step
$$
y=\int_{1}^{x} \sqrt{t^{3}-1} d t , \quad 1 \leq x \leq 4
$$
$\Rightarrow$
$$
y^{\prime} =\sqrt{x^{3}-1}
$$
$\Rightarrow$
$$
\begin{aligned} 1+\left(y^{\prime}\right)^{2}&=1+(\sqrt{x^{3}-1} )^{2} \\
&= x^{3} \\
\end{aligned}
$$
So, the length for the given curve is given by
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{1+\left[y^{\prime}(t)\right]^{2}} d t \\
&=\int_{1}^{4} \sqrt{x^{3}} d x \\
&=\int_{1}^{4} x^{3 / 2} d x \\
&=\frac{2}{5}\left[x^{5 / 2}\right]_{1}^{4} \\
&=\frac{2}{5}(32-1) \\
&=\frac{62}{5} \\
&=12.4
\end{aligned}
$$
