Consider this fallacious argument: If we divide the interval from x= 0 to x= 1 into n equal subintervals, each of length 1/n, we can "approximate" the line from (0, 0) to (1, 1) by a broken line, the horizontal line from (0, 0) to (1/n, 0), the vertical line from I(1/n, 0) to (1/n, 1/n), the horizontal line from (1/n, 1/n) to (2/n, 1/n), the vertical line from (2/n, 1/n) to (2/n, 2/n), etc.
The area under the horizontal line from (k/n, k/n) to ((k+1)/n, k/n) is (k/n)(1/n)= k/n^2 so the total area is \sum_{k=0}^n k/n^2= (1/n^2)\sum_{k=0}^n k= (1/n^2)(n(n+1)/2= (n^2+ n)/2n= n/2+ 1/2. And the limit, as n goes to infinity, is 1/2, the area under the straight line.<br />
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But if we, instead, use the length of the broken line to approximate the length of the line, each segment has length 1/n and there are 2n such segments so that we get a length of <b>2</b> which is wrong. The length of the line segment from (k/n, k/n) to ((k+1)/n, (k+1)/n) simply cannot be approximated by the length of those two horizontal and vertical segments. We have to use the Pythagorean theorem to get the length of each segment: \sqrt{(1/n)^2+ (1/n)^2}= (1/n)\sqrt{2}. And that leads to the arclength formula.
