Exact? Area under a CurveDate: 03/13/98 at 13:50:00 From: Riddle Calculus Class Subject: Exact? area under a curve After much debate and deliberation in class we came up with this question: Would you please explain why the area under a curve is exact and not an approximation? When a fraction has a denominator of infinity, we replace it with zero. Doesn't this fraction just get infinitely small rather than exactly zero? Our example is: the limit of 1/n as n approaches infinity equals zero. On the contrary, 1/n should get close to but never EXACTLY equal zero. Date: 04/06/98 at 10:32:36 From: Doctor Fred Subject: Re: Exact? area under a curve In one sense, you are correct. In your example, 1/n never is actually zero. But the definition of limit gives us that 1/n goes to 0 as n goes to infinity. The same basic idea occurs with areas under curves. Probably the best way to look at it is that either two numbers are the same or they are different. I'll come back to this in a minute. As the number of rectangles used to approximate the area under a curve increases, the sums of those areas gets closer and closer to the actual area. Let me use an actual example to help. If we use inscribed rectangles to approximate the area under x^2 + 1 from 0 to 1, we get that the area in n rectangles is (2n^2 + 3n + 1)/6n^2 = 1 The error is small, and as n gets bigger the error gets smaller. How small is the error? For some n, it is smaller than .01. For some larger n, it is smaller than .000000000000000001. For some larger n, it is smaller than .0000000000000000000000000000000000000000000000000000001 No matter what number you pick, we're closer than that. Suppose then that the numbers (our "estimate" and the exact area) are different. Then their difference is some number, call it K. We can pick some n so big that the difference is smaller than that number K. Then they must not have differed by K. Maybe they differed by some number smaller than K. Then pick a bigger n. If we keep going like that, then it shows our "estimate" is really the same as the exact answer. I hope that helps. -Doctor Fred, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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