Dave's Math Tables: Integral x^n (Math | Calculus | Integrals | Table Of | x^n)

Discoverer: Cavalieri (1598-1647)

Prooved: (n=1,2..9) Cavalieri; (n=positive integer) Fermat (1601-1665) Proof #1: From the derivative

Given :
1. x^m = m x^(m-1)
2. Fundamental Theorem of Calculus m x^(m-1) dx = ../../derivatives/more/x^n.htm x^m dx = x^m + d. (Fundamental Theorem of Calculus) (d = an arbitrary constant) x^(m-1) dx = x^m / m + c (Divide both sides by m) (c=arbitrary constant, d/m = c) x^n dx = x^(n+1) / (n+1) + c (Set m=n+1, substitution) QED. Proof #2: Fermat's Method Known:
1. 1 + r + r^2 + .. + r^n = (1 - r^(n+1)) / (1-r)
2. 1 + r + r^2 + ... = 1 / (1-r) (r < 1) (0 to b) x^n dx is computed by taking the areas of an infinite number of unequall subintervals; larger subintervals at x close to b, smaller when close to 0. (0 to b) f(x) dx = f(b)*(b - br) + f(br)*(br - br^2) + f(br^2)*(br^2 - br^3) + ... (r -> 1-)
= b^n*(b - br) + (br)^n*(br - br^2) + (br^2)^n*(br^2 - br^3) + ...
= b^(n+1)(1-r) + b^(n+1)r^(n+1)(1-r) + b^(n+1)r^(2n+2)(1-r) + ...
= b^(n+1)(1-r) [ 1 + r^(n+1) + (r^(n+1))^2 + ... ]
= b^(n+1)(1-r) [ 1 / (1-r^(n+1)) ] (Theorem 2.)
= b^(n+1) / [ (1 - r^(n+1)) / (1-r) ]
= b^(n+1) / [ 1 + r + r^2 + .. + r^n ] (Theorem 1.)
= b^(n+1) / (n+1) (r -> 1) QED. 