# How do you find the length of the curve y=sqrtx-1/3xsqrtx from x=0 to x=1?

May 24, 2017

$\frac{4}{3}$

#### Explanation:

The length of a curve can be calculated with this integral
${\int}_{a}^{b} = \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Well, we better get deriving! First, take the derivative of $y$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{2}}\right] - \frac{d}{\mathrm{dx}} \left[\frac{1}{3} {x}^{\frac{3}{2}}\right]$

Using power rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left({x}^{- \frac{1}{2}} - {x}^{\frac{1}{2}}\right) = \frac{1 - x}{2 \sqrt{x}}$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \frac{1 - 2 x + {x}^{2}}{4 x}$

So

$\sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} = \sqrt{1 + \frac{1 - 2 x + {x}^{2}}{4 x}} = \sqrt{\frac{1 + 2 x + {x}^{2}}{4 x}} = \frac{x + 1}{2 \sqrt{x}}$

Finally

${\int}_{0}^{1} \frac{x + 1}{2 \sqrt{x}} \mathrm{dx} = {\int}_{0}^{1} \frac{x}{2 \sqrt{x}} + \frac{1}{2 \sqrt{x}} \mathrm{dx}$

Evaluate indefinite

$\int \frac{x}{2 \sqrt{x}} + \frac{1}{2 \sqrt{x}} \mathrm{dx}$
$= \frac{1}{2} \int \frac{x}{\sqrt{x}} + \frac{1}{\sqrt{x}} \mathrm{dx}$

Solve and ignore constants for now

$\int \frac{x}{\sqrt{x}} = \frac{2}{3} {x}^{\frac{3}{2}}$

$\int \frac{1}{\sqrt{x}} = 2 \sqrt{x}$

Piece them together and ignore constants for now

$\frac{1}{2} \int \frac{x}{\sqrt{x}} + \frac{1}{\sqrt{x}} \mathrm{dx}$
$= \frac{1}{2} \left(2 \sqrt{x} + \frac{2}{3} {x}^{\frac{3}{2}}\right)$
$= \frac{\sqrt{x} \left(x + 3\right)}{3}$

Finally, solve the definite integral:
${\int}_{0}^{1} \frac{x + 1}{2 \sqrt{x}} \mathrm{dx} = \frac{\sqrt{x} \left(x + 3\right)}{3} {|}_{0}^{1}$
$= \frac{4}{3}$

$\therefore$ The length of the curve is $\frac{4}{3}$