$\displaystyle [\frac{1}{10x^5}-\frac{1}{6x^3}]$$\displaystyle _{1}^{2}$
The answer is $\displaystyle \frac{779}{240}$
I've tried plugging in 2 for x and 1 for x, but I'm not coming up with the answer.
Thanks
Jason
$\displaystyle [\frac{1}{10x^5}-\frac{1}{6x^3}]$$\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
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?
$\displaystyle y' = \frac{x^3}{2} - \frac{1}{2x^3}$
$\displaystyle (y')^2 = \frac{x^6}{4} - \frac{1}{2} + \frac{1}{4x^6}$
$\displaystyle 1 + (y')^2 = \frac{x^6}{4} + \frac{1}{2} + \frac{1}{4x^6}$
$\displaystyle \sqrt{\frac{x^6}{4} + \frac{1}{2} + \frac{1}{4x^6}} = \frac{x^3}{2} + \frac{1}{2x^3}$
$\displaystyle \int_1^2 \frac{x^3}{2} + \frac{1}{2x^3} \, dx$
$\displaystyle \left[\frac{x^4}{8} - \frac{1}{4x^2}\right]_1^2$
$\displaystyle \left(2 - \frac{1}{16}\right) - \left(\frac{1}{8} - \frac{1}{4}\right) = \frac{33}{16}$