Explanation Given: Let A be the area under y = x − 1 over [ 1 , 8 ] Calculation: Let f ( x ) = x − 1 over the interval [ 1 , 8 ] . Since f is a decreasing function over this interval, the left endpoint approximation L 7 overestimates the true area between the graph f and the x-axis, whereas the right endpoint approximation R 7 underestimates it...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 5.1, Problem 80E

To determine

That the area A under $y = x^{- 1}$ over the interval $[1, 8]$ satisfies $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \leq A \leq 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} .$

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Chapter 5.1, Problem 79E

Chapter 5.1, Problem 81E

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5. mit answer urces Use Simpson's Rule and all the data in the following table to estimate the value of the 31 integral f(x) dx. 25 25 26 27 28 29 30 31 f(x) 4 44 4 -9 -2 9 2 5 (Round your answer to within two decimal places if necessary, but do not round until your final computation.) Simpson's Rule Approximation: PROGRES Score Completi 30 i Submit answer T The Weather Channel UP DELL FB F4 F5 F9 9. F10

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Chapter 5 Solutions

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That the area A under y = x − 1 over the interval [ 1 , 8 ] satisfies 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 ≤ A ≤ 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 .

Solution Summary: The author explains that the area A under y = x 1 over the interval satisfies the given value. The left endpoint approximation L 7 overestimates the true area between the graph

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That the area A under $y = x^{- 1}$ over the interval $[1, 8]$ satisfies $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \leq A \leq 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} .$

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That the area A under y = x − 1 over the interval [ 1 , 8 ] satisfies 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 ≤ A ≤ 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 .

Solution Summary: The author explains that the area A under y = x 1 over the interval satisfies the given value. The left endpoint approximation L 7 overestimates the true area between the graph

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That the area A under y=x−1 over the interval [1, 8] satisfies 12+13+14+15+16+17+18≤A≤1+12+13+14+15+16+17.

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That the area A under $y = x^{- 1}$ over the interval $[1, 8]$ satisfies $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \leq A \leq 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} .$