121616949-math.166 - one that turns out to make sense...

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152 Chapter 7 Integration In other words, the area between the x -axis and the curve, but under the x -axis, “counts as negative area”. So the integral Z 6 0 v ( t ) dt = 18 measures “net area”, the area above the axis minus the (positive) area below the axis. If we recall that the integral is the limit of a certain kind of sum, this behavior is not surprising. Recall the sort of sum involved: n - 1 X i =0 v ( t i t. In each term v ( t t the Δ t is positive, but if v ( t i ) is negative then the term is negative. If over an entire interval, like 5 to 6, the function is always negative, then the entire sum is negative. In terms of area, v ( t t is then a negative height times a positive width, giving a negative rectangle “area”. So now we see that when evaluating Z 6 5 v ( t ) dt = - 17 6 by finding an antiderivative, substituting, and subtracting, we get a surprising answer, but
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Unformatted text preview: one that turns out to make sense. Let’s now try something a bit different: Z 5 6 v ( t ) dt =-t 3 3 + 5 2 t 2 fl fl fl fl 5 6 =-5 3 3 + 5 2 5 2--6 3 3-5 2 6 2 = 17 6 . Here we simply interchanged the limits 5 and 6, so of course when we substitute and subtract we’re subtracting in the opposite order and we end up multiplying the answer by-1. This too makes sense in terms of the underlying sum, though it takes a bit more thought. Recall that in the sum n-1 X i =0 v ( t i )Δ t, the Δ t is the “length” of each little subinterval, but more precisely we could say that Δ t = t i +1-t i , the difference between two endpoints of a subinterval. We have until now...
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