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Exact? Area under a Curve

Date: 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 

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 


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
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Associated Topics:
High School Calculus

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