You can generalise this nicely by using a surface integral instead of a single integral, and by making the substitution u = x^(1/n), y = v^(1/n). Then the Jacobian is n²×u^(n-1)×v^(n-1), and you can integrate with 0 ≤ v ≤ 1 and 0 ≤ u ≤ 1-v. Either way, you get that the nth area is n×Beta(n, n+1).
Yeah, thanks. I initially didn't thought on covering the general version due to it being somewhat advanced, but seeing that this video is gaining popularity, perhaps I should consider it......
A fascinating general expression! However, I don't know if I will cover this topic in near future, because it is bit too advanced compared to the general level of math covered in my channel. Still, thank you for your contribution!
The video about the GENERALIZED version of this problem is now uploaded. Check here: ua-cam.com/video/KAYdQu0FJHc/v-deo.html
You can generalise this nicely by using a surface integral instead of a single integral, and by making the substitution u = x^(1/n), y = v^(1/n). Then the Jacobian is n²×u^(n-1)×v^(n-1), and you can integrate with 0 ≤ v ≤ 1 and 0 ≤ u ≤ 1-v.
Either way, you get that the nth area is n×Beta(n, n+1).
Yeah, thanks. I initially didn't thought on covering the general version due to it being somewhat advanced, but seeing that this video is gaining popularity, perhaps I should consider it......
These can be generalized to Int_0^1(dx(1-x^(1/n))^n)=Gamma(1+n)^2/Gamma(1+2n)
A fascinating general expression! However, I don't know if I will cover this topic in near future, because it is bit too advanced compared to the general level of math covered in my channel. Still, thank you for your contribution!
which is also 1 divided by the nth central binomial coefficient :)
@@blub232324 If n is natural number, then yes!