$\displaystyle [\frac{1}{10x^5}-\frac{1}{6x^3}]$$\displaystyle _{1}^{2}$

The answer is $\displaystyle \frac{779}{240}$

I've tried plugging in 2 for x and 1 for x, but I'm not coming up with the answer.

Thanks

Jason

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- Jul 29th 2009, 12:48 PMDarkhrse99I need help finishing this problem.
$\displaystyle [\frac{1}{10x^5}-\frac{1}{6x^3}]$$\displaystyle _{1}^{2}$

The answer is $\displaystyle \frac{779}{240}$

I've tried plugging in 2 for x and 1 for x, but I'm not coming up with the answer.

Thanks

Jason - Jul 29th 2009, 12:50 PMANDS!
$\displaystyle \int_{1}^{2}[\frac{1/10x^5}{1/6x^3}]

$

Should it be this? - Jul 29th 2009, 01:03 PMDarkhrse99
- Jul 29th 2009, 01:07 PMskeeter
- Jul 29th 2009, 01:14 PMANDS!
Well

$\displaystyle f(x)]_{1}^{2}$

is

$\displaystyle \int_{1}^{2}f(x)dx$

And agreed. That expression evaluated on those bounds can't be the answer you've got; unless this isn't the full problem. - Jul 29th 2009, 01:19 PMDarkhrse99
The original problem is find the arc length of the indicated interval.

$\displaystyle Y=\frac{x^4}{8}+\frac{1}{4x^2}$ - Jul 29th 2009, 01:41 PMANDS!
Whoa. That is markedly different than what you had. What you had simply said "evaluate the integral from 1 to 2". Which is what we're doing when finding arc-length.

So we have an equation of the form

$\displaystyle \int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx$ is equal to the arc-legth of $\displaystyle f(x)$ from $\displaystyle a$ to $\displaystyle b$.

So from your equation we need to find the derivative of:

$\displaystyle y=\frac{x^4}{8}+\frac{1}{4x^2}

$

Which is equal to:

$\displaystyle \frac{x^3}{2}-\frac{1}{2x^{3}}$

Plugging this into our equation for arc length yields:

$\displaystyle \int_{1}^{2}\sqrt{1+[\frac{x^3}{2}-\frac{1}{2x^{3}}]^{2}}dx$

Can you distribute that squared and go from there? - Jul 29th 2009, 01:47 PMskeeter
$\displaystyle y' = \frac{x^3}{2} - \frac{1}{2x^3}$

$\displaystyle (y')^2 = \frac{x^6}{4} - \frac{1}{2} + \frac{1}{4x^6}$

$\displaystyle 1 + (y')^2 = \frac{x^6}{4} + \frac{1}{2} + \frac{1}{4x^6}$

$\displaystyle \sqrt{\frac{x^6}{4} + \frac{1}{2} + \frac{1}{4x^6}} = \frac{x^3}{2} + \frac{1}{2x^3}$

$\displaystyle \int_1^2 \frac{x^3}{2} + \frac{1}{2x^3} \, dx$

$\displaystyle \left[\frac{x^4}{8} - \frac{1}{4x^2}\right]_1^2$

$\displaystyle \left(2 - \frac{1}{16}\right) - \left(\frac{1}{8} - \frac{1}{4}\right) = \frac{33}{16}$ - Jul 29th 2009, 01:54 PMDarkhrse99
Got it guys! Sorry for the confusion.