# Thread: I need help finishing this problem.

1. ## I need help finishing this problem.

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

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

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

Thanks
Jason

2. $\int_{1}^{2}[\frac{1/10x^5}{1/6x^3}]
$

Should it be this?

3. Originally Posted by ANDS!
$\int_{1}^{2}[\frac{1/10x^5}{1/6x^3}]
$

Should it be this?
No, it should be what I said, but the intagral shouldn't be there. I didn't know how to place the 2 and 1 after the brackets. the fraction needs to be evaluated from 1 to 2.

4. Originally Posted by Darkhrse99
No, it should be what I said, but the intagral shouldn't be there. I didn't know how to place the 2 and 1 after the brackets. the fraction needs to be evaluated from 1 to 2.
well, it's not the "answer" you cite (779/240)

maybe it would help if you posted the original definite integral?

5. Well

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

is

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

And agreed. That expression evaluated on those bounds can't be the answer you've got; unless this isn't the full problem.

6. The original problem is find the arc length of the indicated interval.

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

7. 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

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

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

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

Which is equal to:

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

Plugging this into our equation for arc length yields:

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

Can you distribute that squared and go from there?

8. Originally Posted by Darkhrse99
The original problem is find the arc length of the indicated interval.

$Y=\frac{x^4}{8}+\frac{1}{4x^2}$
$y' = \frac{x^3}{2} - \frac{1}{2x^3}$

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

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

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

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

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

$\left(2 - \frac{1}{16}\right) - \left(\frac{1}{8} - \frac{1}{4}\right) = \frac{33}{16}$

9. Got it guys! Sorry for the confusion.