Question

Find the arc length of the curve below on the given interval by integrating with respect to x.

y=2x^(3/2); [0, 1]

Answer 1

The arc length of
`y=2x^(\frac32)` over [0, 1] is given by the integral,

`L=\displaystyle\int_0^1\sqrt{1+\left((\mathrm dy)/(\mathrm dx)\right)^2}\,\mathrm dx`

Differentiate y with respect to x using the power rule:

`(\mathrm dy)/(\mathrm dx)=\frac32*2 x^(\frac32-1)=3x^(\frac12)`

Then the integral becomes

`L=\displaystyle\int_0^1\sqrt{1+\left(3x^(\frac12)\right)^2}\,\mathrm dx=\int_0^1√(1+9x)\,\mathrm dx`

Substitute u = 1 + 9x and du = 9 dx :

`L=\displaystyle\int_0^1√(1+9x)\,\mathrm dx=\frac19\int_1^(10)\sqrt u\,\mathrm du`

`L=\displaystyle\frac19\left(\frac23u^(\frac32)\right)\bigg|_1^(10)`

`L=\frac2{27}\left(10^(\frac32)-1^(\frac32)\right)`

`L=\boxed{\frac2{27}\left(10^(\frac32)-1\right)}`