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Unformatted text preview: , 3] into n equal pieces, of length 31 n = 2 n . The i th piece of this partition has righthand endpoint 1 + i 2 n , so according to the instructions in the problem we have Δ x = 2 n and x * i = 1 + i 2 n . The resulting Riemann sum is: n X i =1 f ( x * i ) · Δ x = n X i =1 ± 1 + i 2 n ² 2 · 2 n . And the integral is the limit of the Riemann sums: Z 3 1 t 2 dt = lim n →∞ n X i =1 ± 1 + i 2 n ² 2 · 2 n . Solution to b): Just plug in n = 4 into our Riemann sum. We get: 4 X i =1 ± 1 + i 2 ² 2 · 1 2 = 1 2 " ± 1 + 1 2 ² 2 + ± 1 + 2 2 ² 2 + ± 1 + 3 2 ² 2 + ± 1 + 4 2 ² 2 # = 1 2 ± 9 4 + 16 4 + 25 4 + 36 4 ² = 43 4 . 2...
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This note was uploaded on 09/17/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at Berkeley.
 Spring '08
 WILKENING
 Math, Calculus, Antiderivatives, Derivative

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