Calculus (Area of a Plane Region)Date: 5/20/96 at 23:21:56 From: Anonymous Subject: Calculus (Area of a Plane Region) The problem is like this: y = 4-x2 ; x axis a) draw a figure showing the region and a rectangular element of area; b) express the area of the region as the limit of a Riemann sum; c) find the limit in part b by evaluating a definite integral by the second fundamental theorem of the calculus. Date: 6/1/96 at 17:54:40 From: Doctor Anthony Subject: Re: Calculus (Area of a Plane Region) The graph of y = 4-x^2 will be an inverted parabola, passing through the points (-2,0), (0,4) and (2,0). We must find the area between this curve and the x axis. By symmetry we can find the area between x=0 (y axis) and x=2, and then double the result. Divide the region by n vertical strips, each of width 2/n. For thin strips we can consider the area of a strip as height times width, with height given by the y value at that position (=4-x^2) and width =2/n. The x value for the rth strip will be r(2/n) where r takes all values from 1 to n. By summing all these areas, we obtain the area under the curve, provided n is taken sufficiently large. Area under curve = SIGMA[4-4r^2/n^2](2/n) for r = 1 to n = (8/n)SIGMA[1 - r^2/n^2] for r = 1 to n = (8/n)[n - (1/n^2){n(n+1)(2n+1)/6}] = (8/n)[n - (1/6n^2){2n^3+3n^2+n}] = (8/n)[n - (1/6){2n+ 3 + 1/n}] = (8/n)[n - (1/3)n + 1/2 + 1/6n] = (8/n)[(2/3)n + 1/2 + 1/6n] = (16/3) + 4/n + 4/3n^2 Now let n -> infinity and the last two terms tend to zero, so the area is 16/3. The total area required is twice this value, i.e. 32/3 If we do normal integration of 4-x^2 between -2 an +2 we have INT[(4-x^2)dx] from -2 to +2 = [4x-x^3/3] from -2 to +2 = {8-8/3}-{-8+8/3} = 8 - 8/3 + 8 - 8/3 = 32/3 And so this illustrates the theory that the integral gives the area under the curve. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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