# How do you find the length of the curve for y=2x^(3/2) for (0, 4)?

Feb 7, 2016

$S = \frac{1}{6} \left({\left(37\right)}^{\frac{3}{2}} - 1\right)$

#### Explanation:

Given equation is $f \left(x\right) = 2 {x}^{\frac{3}{2}}$.
We are given the task to find the length of the curve of the given equation in the interval $\left(0 , 4\right)$.

The equation to find length of the curve is $S = \setminus {\int}_{0}^{4} \setminus \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$
So, the derivative of the given equation will be
$f ' \left(x\right) = 2 \frac{d}{\mathrm{dx}} \left({x}^{\frac{3}{2}}\right) = 2 \cdot \frac{3}{2} \cdot {x}^{\frac{3}{2} - 1} = 3 \cdot {x}^{\frac{1}{2}}$

So substituting for $f ' \left(x\right)$, $S = \setminus {\int}_{0}^{4} \setminus \sqrt{1 + 9 x} \mathrm{dx}$
Taking $9 x = t \setminus \implies \mathrm{dx} = \frac{\mathrm{dt}}{9}$ and that at $x = 0 \setminus \implies t = 0$ and $x = 4 \setminus \implies t = 36$
So the given integral becomes $S = \frac{1}{9} \setminus {\int}_{0}^{36} \setminus \sqrt{1 + t} \mathrm{dt}$
So, $S = {\cancel{3}}^{1} / 2 \cdot \frac{1}{\setminus} {\cancel{9}}^{3} {\left(1 + t\right)}^{\frac{3}{2}} {|}_{0}^{36}$
Applying limits and totalling, I get the above answer.