Integral Problem Help

**Simon777** · February 21st 2010, 07:05 AM

The problem is y=x^(2/3) and x=1 and x=8. I need to find the exact arc length of the curve over the interval.

The formula used is the arc length formula: L=integral from a to b of (1+(dy/dx)^2) ^(1/2)

So far I got this:

Integral from 1 to 8 of (1+(4/9*x^(-2/3)))^(1/2)dx

I don't know how to integrate this. Is it a u substitution? I tried u= 1+ 4/9x^(-2/3) and I have an x^(-5/3) left over so that won't work.

**Ted** · February 21st 2010, 07:24 AM

Originally Posted by Simon777

The problem is y=x^(2/3) and x=1 and x=8. I need to find the exact arc length of the curve over the interval.

The formula used is the arc length formula: L=integral from a to b of (1+(dy/dx)^2) ^(1/2)

So far I got this:

Integral from 1 to 8 of (1+(4/9*x^(-2/3)))^(1/2)dx

I don't know how to integrate this. Is it a u substitution? I tried u= 1+ 4/9x^(-2/3) and I have an x^(-5/3) left over so that won't work.

For:
$\int_1^8 \sqrt{\frac{4}{9x^{\frac{2}{3}}}+1} dx$ .

Make a common denominator to get:

$\int_1^8 \sqrt{ \frac{ 9x^{\frac{2}{3}}+4 }{ 9x^{\frac{2}{3} }} }dx$ .

Now, Apply : $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ for $b \neq 0$ .

Then, think about a substitution.

**skeeter** · February 21st 2010, 07:30 AM

Originally Posted by Simon777

The problem is y=x^(2/3) and x=1 and x=8. I need to find the exact arc length of the curve over the interval.

The formula used is the arc length formula: L=integral from a to b of (1+(dy/dx)^2) ^(1/2)

So far I got this:

Integral from 1 to 8 of (1+(4/9*x^(-2/3)))^(1/2)dx

I don't know how to integrate this. Is it a u substitution? I tried u= 1+ 4/9x^(-2/3) and I have an x^(-5/3) left over so that won't work.

$f'(x) = \frac{2}{3}x^{-\frac{1}{3}}$

$f'(x) = \frac{2}{3x^{\frac{1}{3}}}$

$[f'(x)]^2 = \frac{4}{9x^{\frac{2}{3}}}$

$1 + [f'(x)]^2 = 1 + \frac{4}{9x^{\frac{2}{3}}} = \frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}<br />$

$\sqrt{\frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}} = \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4}$

$\int_1^8 \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4} \, dx$

let $u = 9x^{\frac{2}{3}}+4$

$du = \frac{6}{x^{\frac{1}{3}}} \, dx$

$\frac{1}{18} \int_{13}^{40} \sqrt{u} \, du$

can you finish ?

**Simon777** · February 21st 2010, 07:32 AM

Originally Posted by Ted

For:
$\int_1^8 \sqrt{\frac{4}{9x^{\frac{2}{3}}}+1} dx$ .

Make a common denominator to get:

$\int_1^8 \sqrt{ \frac{ 9x^{\frac{2}{3}}+4 }{ 9x^{\frac{2}{3} }} }dx$ .

Now, Apply : $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ for $b \neq 0$ .

Then, think about a substitution.

Wow, such a simple solution, I don't know why I didn't think to do that. Thank you so much for your help.

**Simon777** · February 21st 2010, 07:35 AM

Originally Posted by skeeter

$f'(x) = \frac{2}{3}x^{-\frac{1}{3}}$

$f'(x) = \frac{2}{3x^{\frac{1}{3}}}$

$[f'(x)]^2 = \frac{4}{9x^{\frac{2}{3}}}$

$1 + [f'(x)]^2 = 1 + \frac{4}{9x^{\frac{2}{3}}} = \frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}<br />$

$\sqrt{\frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}} = \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4}$

$\int_1^8 \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4} \, dx$

let $u = 9x^{\frac{2}{3}}+4$

$du = \frac{6}{x^{\frac{1}{3}}} \, dx$

$\frac{1}{18} \int_{13}^{40} \sqrt{u} \, du$

can you finish ?

Thank you for working it out so I have something to check my steps with. Yes, I finished and checked my answer with the back of my book and it is correct.

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