121616949-math.165 - 5 v t dt Computing the two integrals...

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7.3 Some Properties of Integrals 151 - 6 - 4 - 2 0 2 4 6 1 2 3 4 5 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ......................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 5 10 15 20 1 2 3 4 5 6 ................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................................................. . . . . Figure 7.2 The velocity of an object and its position. As we have seen, we can also compute distance traveled with an integral; let’s try it. Z 6 0 v ( t ) dt = Z 6 0 - t 2 + 5 t dt = - t 3 3 + 5 2 t 2 6 0 = 18 . What went wrong? Well, nothing really, except that it’s not really true after all that “we can also compute distance traveled with an integral”. Instead, as you might guess from this example, the integral actually computes the net distance traveled, that is, the difference between the starting and ending point. As we have already seen, Z 6 0 v ( t ) dt = Z 5 0 v ( t ) dt + Z 6
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Unformatted text preview: 5 v ( t ) dt. Computing the two integrals on the right (do it!) gives 125 / 6 and-17 / 6, and the sum of these is indeed 18. But what does that negative sign mean? It means precisely what you might think: it means that the object moves backwards. To get the total distance traveled we can add 125 / 6 + 17 / 6 = 71 / 3, the same answer we got before. Remember that we can also interpret an integral as measuring an area, but now we see that this too is a little more complicated that we have suspected. The area under the curve v ( t ) from 0 to 5 is given by Z 5 v ( t ) dt = 125 6 , and the “area” from 5 to 6 is Z 6 5 v ( t ) dt =-17 6 ....
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