121616949-math.155 - 3 t 2 n 2(0 1 2 ·· n-1 = 3 t 2 n 2...

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7.1 Two examples 141 that is, 3 /n 2 times the sum of the first n - 1 positive integers. Now we make use of a fact you may have run across before: 1 + 2 + 3 + · · · + k = k ( k + 1) 2 . In our case we’re interested in k = n - 1, so 1 + 2 + 3 + · · · + ( n - 1) = ( n - 1)( n ) 2 = n 2 - n 2 . This simplifies the approximate distance traveled to 3 n 2 n 2 - n 2 = 3 2 n 2 - n n 2 = 3 2 n 2 n 2 - n n 2 = 3 2 1 - 1 n . Now this is quite easy to understand: as n gets larger and larger this approximation gets closer and closer to (3 / 2)(1 - 0) = 3 / 2, so that 3 / 2 is the exact distance traveled during one second, and the final position is 11 . 5. So for t = 1, at least, this rather cumbersome approach gives the same answer as the first approach. But really there’s nothing special about t = 1; let’s just call it t instead. In this case the approximate distance traveled during time interval number i is 3( i - 1)( t/n )( t/n ) = 3( i - 1) t 2 /n 2 and the total distance traveled is approximately (0) t n + 3(1) t 2 n 2 + 3(2) t 2 n 2 + 3(3) t 2 n 2 + · · · + 3( n - 1) t 2 n 2
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Unformatted text preview: 3 t 2 n 2 (0 + 1 + 2 + ··· + ( n-1)) = 3 t 2 n 2 n 2-n 2 = 3 2 t 2 ± 1-1 n ¶ . In the limit, as n gets larger, this gets closer and closer to (3 / 2) t 2 and the approximated position of the object gets closer and closer to (3 / 2) t 2 + 10, so the actual position is (3 / 2) t 2 + 10, exactly the answer given by the first approach to the problem. EXAMPLE 7.2 Find the area under the curve y = 3 x between x = 0 and any positive value x . There is here no obvious analogue to the first approach in the previous example, but the second approach works fine. (Because the function y = 3 x is so simple, there is another approach that works here, but it is even more limited in potential application than is approach number one.) How might we approximate the desired area? We know...
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