The Area Under a Parabola and the Length of a Curve
Date: 11/15/95 at 21:57:28 From: Navigator Subject: area under parabola The following problems are an enigma to me. Find the area under the parabola f(x) = x^2 + 1 for x = 0 to x = 2, and y > 0. Also find the length of the curve from x = 0 to x = 2. Find the area under the parabola f(x) = x^2 - 2x + 4 and the horizontal axis for x = 0 to x = 3, and y > 0. Also, find the length of the curve from x = 0 to x = 3. I need help on these problems please. John email@example.com
Date: 11/19/95 at 17:15:20 From: Doctor Josh Subject: Re: area under parabola Dear John, The area under the first parabola is the integral from 0 to 2 of x^2+1. Thus, / Area = |[the integral sign] (x^2 + 1)dx from 0 to 2 / = [(1/3)(2^3) + 2] - [(1/3)(0^3) + 0] = 14/3 square units The length of the curve from x=0 to x=2 is: / length = |[1 - (y')^2]dx from x=0 to x=2 / Draw out a triangle of dx, dy, and dl to see why. Use the same techniques for the other parabola. -Doctor Josh, The Geometry Forum
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