121616949-math.154 - 140 Chapter 7 Integration look at the...

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140 Chapter 7 Integration look at the times 0 . 1, 0 . 2, 0 . 3, . . . , 1 . 0, and try approximating the location of the object at each, by supposing that during each tenth of a second the object is going at a constant speed. Since the object initially has speed 0, we again suppose it maintains this speed, but only for a tenth of second; during that time the object would not move. During the tenth of a second from t = 0 . 1 to t = 0 . 2, we suppose that the object is traveling at 0 . 3 cm/sec, namely, its actual speed at t = 0 . 1. In this case the object would travel (0 . 3)(0 . 1) = 0 . 03 centimeters: 0 . 3 cm/sec times 0 . 1 seconds. Similarly, between t = 0 . 2 and t = 0 . 3 the object would travel (0 . 6)(0 . 1) = 0 . 06 centimeters. Continuing, we get as an approximation that the object travels (0 . 0)(0 . 1) + (0 . 3)(0 . 1) + (0 . 6)(0 . 1) + · · · + (2 . 7)(0 . 1) = 1 . 35 centimeters, ending up at position 11.35. This is a better approximation than 10, certainly, but is still just an approximation. (We know in fact that the object ends up at position 11 . 5, because we’ve already done the problem using the first approach.)
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