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.