- #1

- 29

- 0

[tex]\int^{6}_{0} \sqrt{1-n^2x^2}dx=\pi+e[/tex]

I need to solve this for n. I believe there should only be one possible function of the form [tex]y=x^n[/tex] that gives an arclength of [tex]\pi+e[/tex] over the interval x=0 to x=6, and wish to find the value of n that such a function must have.

Does anyone know how to do this? I haven't the slightest idea, as I only know as much calculus as I've managed to teach myself over the past few months... Thank you!

Ah, additionally, I'm assuming (as I, regrettably, read somewhere) that

[tex]\int^{b}_{a} \sqrt{1-[f'(x)]^2}dx[/tex]

is equal to arclength (actually, I didn't just accept it completely--I lack the mathematics to evaluate whether or not it actually is such a formula, but my TI-89 is capable of calculating for whatever values I plug in so... They have thus far matched up perfectly with the values produced by the method I came up with myself:)

[tex]\lim_{x \rightarrow 0}\sum^{\frac{m}{x}-1}_{n=0}\sqrt{x^2+(f(x(n+1))-f(nx))^2}[/tex]

Anyway, again, thank you.

I need to solve this for n. I believe there should only be one possible function of the form [tex]y=x^n[/tex] that gives an arclength of [tex]\pi+e[/tex] over the interval x=0 to x=6, and wish to find the value of n that such a function must have.

Does anyone know how to do this? I haven't the slightest idea, as I only know as much calculus as I've managed to teach myself over the past few months... Thank you!

Ah, additionally, I'm assuming (as I, regrettably, read somewhere) that

[tex]\int^{b}_{a} \sqrt{1-[f'(x)]^2}dx[/tex]

is equal to arclength (actually, I didn't just accept it completely--I lack the mathematics to evaluate whether or not it actually is such a formula, but my TI-89 is capable of calculating for whatever values I plug in so... They have thus far matched up perfectly with the values produced by the method I came up with myself:)

[tex]\lim_{x \rightarrow 0}\sum^{\frac{m}{x}-1}_{n=0}\sqrt{x^2+(f(x(n+1))-f(nx))^2}[/tex]

Anyway, again, thank you.

Last edited: