Lesson22_-_Areas_and_Distances_ws-sol

Lesson22_-_Areas_and_Distances_ws-sol - Find lim n →∞ L...

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Solutions to Worksheet for Section 5.1 Areas and Distances V63.0121, Calculus I Spring 2010 1. Draw the graph of f ( x ) = e x on the interval [0 , 1] . We are going to find the area below the curve on this interval. Solution. Here is the region: x y 1 1 2 3 2. Estimate the area by computing with a calculator. (i) L 2 and R 2 (ii) L 4 and R 4 (iii) L 8 and R 8 Solution. We have L 2 = e 0 (0 . 5) + e 0 . 5 (0 . 5) = 1 . 32436 R 2 = e 0 . 5 (0 . 5) + e 1 . 0 (0 . 5) = 2 . 18350 And so on. n R n L n 2 2.18350 1.32436 4 1.94201 1.51244 8 1.82791 1.61313 1
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3. Show that L n = 1 n ± 1 + e 1 /n + e 2 /n + · · · + e ( n - 1) /n ² = n X i =1 e ( i - 1) /n n R n = 1 n ± e 1 /n + e 2 /n + · · · + e ² = n X i =1 e i/n n . 4. Find “closed form” expressions (i.e., without ellipses or sigmas) for L n and R n . Hint. Remember that for any number r that 1 + r + r 2 + r 3 + · · · + r n - 1 = 1 - r n 1 - r . Solution. Using the formula from on L n with r = e 1 /n , we have L n = 1 n n X i =1 ± e 1 /n ² i - 1 = 1 - e n (1 - e 1 /n ) If you look carefully, you notice R n = e 1 /n L n , so R n = e 1 /n (1 - e ) n (1 - e 1 /n ) 5.
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Unformatted text preview: Find lim n →∞ L n and lim n →∞ R n . These limits should be the same, and their common value is the area of the region. Solution. Following the hint, lim n →∞ L n = lim n →∞ 1-e n (1-e 1 /n ) Write h = 1 /n . So lim n →∞ L n = lim n →∞ 1-e n (1-e 1 /n ) = (1-e ) lim h → + h 1-e h = (1-e )(-1) ³ lim h → + e h-1 h ´-1 The limit in the parentheses is the derivative of f ( x ) = e x at a = 0. We know that limit is 1. Hence lim n →∞ L n = e-1 Since R n = e 1 /n L n , we have lim n →∞ R n = ± lim n →∞ e 1 /n ²± lim n →∞ L n ² = 1 · ( e-1) 2...
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This note was uploaded on 07/19/2010 for the course MATHEMATIC V63.0121 taught by Professor Leingang during the Spring '09 term at NYU.

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Lesson22_-_Areas_and_Distances_ws-sol - Find lim n →∞ L...

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