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 26 3 35 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 by1. This too makes sense in terms of the underlying sum, though it takes a bit more thought. Recall that in the sum n1 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 +1t i , the diﬀerence between two endpoints of a subinterval. We have until now...
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 Spring '07
 JonathanRogawski
 Math, Calculus, Addition, negative height times

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