what is integral of Mx?
Answers
Answer:
x^m = m x^(m-1)
Fundamental Theorem of Calculus
(integral)m x^(m-1) dx = (integral)../../derivatives/more/x^n.htm (d-dx)x^m dx = x^m + d. (Fundamental Theorem of Calculus) (d = an arbitrary constant)
(integral)x^(m-1) dx = x^m / m + c (Divide both sides by m) (c=arbitrary constant, d/m = c)
(integral)x^n dx = x^(n+1) / (n+1) + c (Set m=n+1, substitution) QED.
Proof #2: Fermat's Method
Known:
1 + r + r^2 + .. + r^n = (1 - r^(n+1)) / (1-r)
1 + r + r^2 + ... = 1 / (1-r) (r < 1)
(integral)(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.
(integral)(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.
Step-by-step explanation:
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