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Unformatted text preview: M 408L  Integral Calculus Worksheet  Homework 3 February 2, 2009 Group 1 : Problem like #3. Consider the following Riemann sum: 2 n 4 + 2 n 2 + 2 n 4 + 4 n 2 + 2 n 4 + 6 n 2 + ··· + 2 n 4 + 2 n n 2 1. Rewrite the sum using sigma notation starting with i = 1. How does your answer change if the sum starts at i = 7? Solution: We want to write the above sum, using sigma notation, as ∑ n i =1 f ( x * i )Δ x . Therefore, we’ll have ∑ n i =1 2 n ( 4 + 2 i n ) 2 . We now want to write the same sum again using sigma notation, but this time we want to start with i = 7. Remember, we don’t want the things we add up to change, we just want to start the counter in a different place. So when i = 7 we want to have 2 n ( 4 + 2 · 1 n ) 2 . When i = 8 we want 2 n ( 4 + 2 · 2 n ) 2 , and when i = 9 we want 2 n ( 4 + 2 · 3 n ) 2 . Therefore, our sum should be ? X i =7 2 n 4 + 2( i 6) n . Notice that we have not yet found where we should stop. For i =?, we want to have 2 n ( 4 + 2 n n ) 2 . Therefore, ? 6 = n , so ? = n 6 and our sum is as follows: n 6 X i =7 2 n 4 + 2( i 6) n . 2. If the above sum is the n th Riemann sum approximation for a definite integral, what is the integral? Solution: In order to do this problem we will have to make a few choices....
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 Fall '09
 RAdin
 Calculus, Riemann sum, Riemann, ln tdt

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