Expected Value of Area of TriangleDate: 05/10/2000 at 10:16:17 From: Aled Jones Subject: Uniform Distribution AB is the diameter of a circle, with radius r. AP is a chord. X is the angle between AB and AP. X is uniformly distributed over the interval 0 to pi/2. Find the expected value of the area of triangle APB. Date: 05/10/2000 at 14:59:55 From: Doctor Anthony Subject: Re: Uniform Distribution The area of triangle APB = (1/2)2r.2r.cos(x).sin(x) = 2r^2.cos(x).sin(x) = r^2.sin(2x) The expected value of the area is then INT(0 to pi/2)[r^2.sin(2x).dx/(pi/2)] = (2/pi)r^2.(-1/2)cos(2x) from 0 to pi/2 = (-1/pi)r^2[cos(pi) - cos(0)] = (-1/pi)r^2[-1 - 1] = 2.r^2/pi - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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