121616949-math.207 - missing is Δ t —but in fact Δ t =...

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9.4 Average value of a function 193 yet we can’t merely add up some number of speeds and divide, since the speed is changing continuously over the time interval. To make sense of “average” in this context, we fall back on the idea of approximation. Consider the speed of the object at tenth of a second intervals: sin 0, sin(0 . 1 π ), sin(0 . 2 π ), sin(0 . 3 π ), . . . , sin(0 . 9 π ). The average speed “should” be fairly close to the average of these ten speeds: 1 10 9 X i =0 sin( πi/ 10) 1 10 6 . 3 = 0 . 63 . Of course, if we compute more speeds at more times, the average of these speeds should be closer to the “real” average. If we take the average of n speeds at evenly spaced times, we get: 1 n n - 1 X i =0 sin( πi/n ) . Here the individual times are t i = i/n , so rewriting slightly we have 1 n n - 1 X i =0 sin( πt i ) . This is almost the sort of sum that we know turns into an integral; what’s apparently
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Unformatted text preview: missing is Δ t —but in fact, Δ t = 1 /n , the length of each subinterval. So rewriting again: n-1 X i =0 sin( πt i ) 1 n = n-1 X i =0 sin( πt i )Δ t. Now this has exactly the right form, so that in the limit we get average speed = Z 1 sin( πt ) dt =-cos( πt ) π fl fl fl fl 1 =-cos( π ) π + cos(0) π = 2 π ≈ . 6366 ≈ . 64 . It’s not entirely obvious from this one simple example how to compute such an average in general. Let’s look at a somewhat more complicated case. Suppose that the velocity of an object is 16 t 2 + 5 feet per second. What is the average velocity between t = 1 and t = 3? Again we set up an approximation to the average: 1 n n-1 X i =0 16 t 2 i + 5 ,...
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