Evaluating integral (arclength)

[TsAmE]

Calculate the length of the curve $\displaystyle y = 3(\sqrt[3]{x^{2}}-1)$ from B(1,0) to C(8,9)

I got stuck with the integral: $\displaystyle \int_{1}^{8} \sqrt{1 + 4x^{\frac{-2}{3}}} dx$ and dont know how you get to $\displaystyle \int_{1}^{8}\frac{\sqrt{x^{\frac{2}{3}}+4}}{|x^{\frac{1}{3}}|}dx$ (which was the next step of the solution).

 

[Deleted member 4993]

Calculate the length of the curve $\displaystyle y = 3(\sqrt[3]{x^{2}}-1)$ from B(1,0) to C(8,9)

I got stuck with the integral: $\displaystyle \int_{1}^{8} \sqrt{1 + 4x^{\frac{-2}{3}}} dx$ and dont know how you get to $\displaystyle \int_{1}^{8}\frac{\sqrt{x^{\frac{2}{3}}+4}}{|x^{\frac{1}{3}}|}dx$ (which was the next step of the solution).

Just multiply by 1 - that is multiply by:

$\displaystyle \frac{\sqrt{x^{\frac{2}{3}}}}{|x^{\frac{1}{3}}|}$

Try substitution:

tan(?) = 2x[sup:24ferbux](1/3)[/sup:24ferbux]

 

[soroban]

Hello, TsAmE!

\(\displaystyle \text}Calculate the length of the curve }\,y \:=\: 3\left(\sqrt[3]{x^{2}}-1\right)\:\text{ from }B(1,0)\text{ to }C(8,9)\)

$\displaystyle \text{I got stuck with the integral: }\:\int_{1}^{8} \sqrt{1 + 4x^{-\frac{2}{3}}}\, dx$

$\displaystyle \text{The integrand is: }\;\sqrt{1 + \frac{4}{x^{\frac{2}{3}}}} \;=\;\sqrt{\frac{x^{\frac{2}{3}} + 4}{x^{\frac{2}{3}}}} \;=\;\frac{\sqrt{x^{\frac{2}{3}}+4}}{x^{\frac{1}{3}}}$


\(\displaystyle \text{We have: }\;L \;=\;\int^8_1\frac{\sqrt{x^{\frac{2}{3}}+4}}{x^{\frac{1}{3}}}\,dx \;=\;\int^8_1\left(x^{\frac{2}{3}} + 4)^{\frac{1}{2}}\lefet(x^{-\frac{1}{3}}dx\right)\)


$\displaystyle \text{Let: }\,u \:=\:x^{\frac{2}{3}}+4 \quad\Rightarrow\quad du \:=\:\tfrac{2}{3}x^{-\frac{1}{3}}dx \quad\Rightarrow\quad x^{-\frac{1}{3}}dx \:=\:\tfrac{3}{2}du$


$\displaystyle \text{Substitute: }\;L \;=\;\int u^{\frac{1}{2}}\left(\tfrac{3}{2}du\right) \;=\;\tfrac{3}{2}\int u^{\frac{1}{2}}\,du \;=\;u^{\frac{3}{2}} + C$


$\displaystyle \text{Back-substitute: }\;L \;=\; \left(x^{\frac{2}{3}}+4\right)^{\frac{3}{2}}\,\bigg]^8_1$


\(\displaystyle \text{Evaluate: }\;L \;\;=\;\;\left(8^{\frac{2}{3}} + 4\right)^\frac{3}{2}} - \left(1^{\frac{2}{3}} + 4\right)^{\frac{3}{2}}\right) \;\;=\;\;(4+4)^{\frac{3}{2}} - (1+4)^{\frac{3}{2}}\)

. . . . . . . . . .$\displaystyle =\;\; 8^{\frac{3}{2}} - 5^{\frac{3}{2}} \;\;=\;\; 8\sqrt{8} - 5\sqrt{5} \;\;=\;\;16\sqrt{2} - 5\sqrt{5}$