Question

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Answer 1

`y=\displaystyle\int_1^x√(t^3-1)\,\mathrm dt`

By the fundamental theorem of calculus,


`(\mathrm dy)/(\mathrm dx)=(\mathrm d)/(\mathrm dx)\displaystyle\int_1^x√(t^3-1)\,\mathrm dt=√(x^3-1)`

Now the arc length over an arbitrary interval
`(a,b)` is


`\displaystyle\int_a^b\sqrt{1+\left((\mathrm dy)/(\mathrm dx)\right)^2}\,\mathrm dx=\int_a^b√(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<b`. 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)`