\(y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt\)
By the fundamental theorem of calculus,
\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}\)
Now the arc length over an arbitrary interval \((a,b)\) is
\(\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx\)
But before we compute the integral, first we need to make sure the integrand exists over it. \(x^{3/2}\) is undefined if \(x<0\), so we assume \(a\ge0\) and for convenience that \(a. Then
\(\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)\)
