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# ma019 - The Evolution of Integration A Shenitzer and J...

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The Evolution of Integration A. Shenitzer and J. Stepr¯ans The American Mathematical Monthly , January 1994, Volume 101, Number 1, pp.66–72. T he Greek Period. The Greek problem underlying integration is the quadrature problem: Given a plane figure, construct a square of equal area. It is easy to solve the quadrature problem for a polygon, a figure with rectilinear boundary. The first quadrature of a figure with curvilinear boundary was achieved by Hippocrates in the fifth century B.C. Hippocrates showed that the area of the lunule in Figure 1 (that is, the figure bounded by one-half of a circle of radius 1 and one-quarter of a circle of radius is equal to the area of the unit square B . Hippocrates managed to square two other lunules.* In the third century B.C. Archimedes effected the quadrature of a parabolic segment. He showed that its area is where is the triangle of maximal area inscribed in the parabolic segment. Archimedes effected a number of other quadratures (and cubatures). Some of his quadratures involved inventive constructions but most relied on the technique of wedging an area between ever closer upper and lower approximating sums. Figure 1 Figure 2 Analogs of such sums are a key element of the definition of the Darboux integral (a variant of the Riemann integral introduced by Darboux in the 19th century) as well as of quadrature programs for computers. We illustrate both of Archimedes’ approaches next. *Two more quadrable lunules were found by T. Clausen in the 19th century. In the 20th century, two Russian algebraists proved (independently) that these five lunules are the only quadrable ones. D 4 3 D , ! 2 d

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Figure 3 Consider Figure 3. Here the hypotenuse AB of the right triangle OAB is tangent to the spiral at A . It then turns out that the side AB is equal to the circumference of the circle with radius OA . (This is a special case of Archimedes’ rectification of circular arcs by using tangents to spirals.) Since he knew that the area of a circle is half the product of its circumference by its radius, we can say that Archimedes used (a tangent to) a spiral to rectify a circle and square its area. Their brilliance notwithstanding, such constructions have been reduced to historical footnotes because they failed to yield general methods. Figure 4 shows a turn of Archimedes’ spiral and the associated circle of radius 2 and thus of area To compute the area S of the turn of the spiral in Figure 4 Archimedes approximates it from below and above by unions of circular sectors indicated in Figure 5. K 5 4 p 3 a 2 . p a , r 5 a u 2
Figure 4 Figure 5 The areas of these approximating figures are, respectively, and It is not difficult to see that for all n . This double inequality can be rewritten as for all n . Obviously, for all n . To prove that Archimedes shows that and is thus small for large n . He can now show that the assumption leads to a contradiction and can conclude that While Archimedes makes no explicit use of limits, he relies on the “method of

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ma019 - The Evolution of Integration A Shenitzer and J...

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