Area Under a ParabolaDate: 06/27/97 at 08:16:19 From: George Osbert Matthew Subject: Calculus Why is the area of a parabolic segment equal to b^3/3? Date: 06/27/97 at 10:04:44 From: Doctor Rob Subject: Re: Calculus This formula is the area in the first quadrant (x >= 0, y >= 0) bounded by the line x = b and the parabola y = x^2. It is found by evaluating the definite integral: _ b _ b |b A = / y dx = / x^2 dx = x^3/3| = b^3/3 _/ 0 _/ 0 |x=0 If your question is why the area is given by this integral, the answer is that the integral is the limit of a sum of areas of rectangles under the curve whose base has length delta x and whose height from the axis to the curve is x^2, where the limit is taken as delta x approaches zero. Each of these sums is an approximation to the area, and since the limit exists, its value is taken as the area in question. If your question is why the limit of this sum has the value b^3/3, this comes from looking at the sum carefully. Let delta x = b/n, for some integer n. Then each rectangle will have base b/n and height (i*b/n)^2, for i = 0, 1, ..., n-1. All of these rectangles are inside the area we are considering, so the sum of all their areas, A(n), satisfies: n-1 n-1 A > A(n) = Sum (b/n)(i*b/n)^2 = (b/n)^3 Sum i^2 i=0 i=0 = (b/n)^3*n*(n-1)*(2*n-1)/6 = b^3/3*(1-1/n)*(1-1/(2*n)) Clearly the limit of this expression as n goes to infinity is b^3/3. Similarly, if we take rectangles which collectively include the area A, then they have base b/n and height (i*b/n)^2, where i = 1, 2, ., n. The sum of their areas, B(n), satisfies: n n A < B(n) = Sum (b/n)(i*b/n)^2 = (b/n)^3 Sum i^2 i=1 i=1 = (b/n)^3*n*(n+1)*(2*n+1)/6 = b^3/3*(1+1/n)*(1+1/(2*n)) Clearly the limit of this expression as n goes to infinity is also b^3/3. Since A(n) < A < B(n), and lim A(n) = b^3/3 = lim B(n), it must be that A = b^3/3. If I have not answered your question, please write again with a fuller explanation of what you need. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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