Finding Function with Definite Integral & Arc Length

[bpcraig]

Sorry, latex is being weird.

I'm currently trying to come up with a way to find an equation that satisfies:

$$s=\int_a^b \sqrt{(f'[x])^2+1} \, dx$$

Which is arc length

and

$$G=\int_a^b f[x] \, dx$$

which is area under the curve

where A and s are known values, and f[a]=A, f=B

I've tried expressing f[x] as a power series, namely:

$$f[x]=\sum _{n=0}^N a_nx^n$$

so that:

$$f'[x]=\sum _{n=1}^N n x^{-1+n} a_n$$

But using that in

$$s=\int_a^b \sqrt{(f'[x])^2+1} \, dx$$

has proven to be difficult as I am unclear how to square a power series, let alone how to integrate it's root.

I've also tried adapting the power series into a finite product, namely:

$$\text{Log}\left[\sum _{n=0}^N e^{\left(a_nx^n\right)}\right]$$

Where f'[x] is:

$$f'[x]=\text{Log}\left[\sum _{n=1}^N e^{\left(n a_nx^{n-1}\right)}\right]$$

But I encounter a similar problem, for some r, ln[r]^2 can not be simplified.

any insight would be greatly appreciated, thanks!

 

[jvicens]

Dear fellow scientist,

I understand your frustration with latex acting up, but let's focus on the problem at hand. Finding an equation that satisfies both arc length and area under the curve is an interesting challenge.

One approach you could try is using the Fundamental Theorem of Calculus. This states that the derivative of the integral of a function is the original function. So for your first equation, you could start by taking the derivative of s with respect to x, which would give you:

s'=\sqrt{(f'[x])^2+1}

Then, using the Fundamental Theorem of Calculus, you could integrate both sides with respect to x to get:

s=\int_a^b \sqrt{(f'[x])^2+1} \, dx

This might make it easier to work with the square root and the power series.

For the second equation, you could try using the fact that the area under a curve is equal to the integral of the function. So you could start by setting G equal to the integral of f[x]:

G=\int_a^b f[x] \, dx

And then use the power series for f[x] to integrate it and solve for G.

I hope this helps and gives you some new ideas to try. Good luck with your research!