Explanation Given information: g ( t ) = 1 + t and the interval is 0 ≤ t ≤ 2 . Concept used: For the interval a ≤ t ≤ b average value of function f ( t ) is f a v e = 1 b − a ∫ a b f ( t ) d t .......... ( 1 ) Calculation: Here the function is. g ( t ) = 1 + t And the interval is in between t = 0 and t = 2 . Now using ( 1 ) , the average value can be calculated as follows: g a v e = 1 2 − 0 ∫ 0 2 ( 1 +

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 5.6, Problem 3P

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Find the average value of the function over the interval 0≤t≤2.

g(t)=1+t for 0≤t≤2

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Chapter 5.6, Problem 2P

Chapter 5.6, Problem 4P

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Chapter 5 Solutions

Applied Calculus, 6e WileyPLUS + Loose-leaf

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Find the average value of the function over the interval 0 ≤ t ≤ 2 . g ( t ) = 1 + t for 0 ≤ t ≤ 2

Solution Summary: The author calculates the average value of the function using g(t)=1+t.

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Find the average value of the function over the interval 0≤t≤2.

g(t)=1+t for 0≤t≤2

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Chapter 5 Solutions

Find the average value of the function over the interval 0 ≤ t ≤ 2 . g ( t ) = 1 + t for 0 ≤ t ≤ 2

Solution Summary: The author calculates the average value of the function using g(t)=1+t.

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Find the average value of the function over the interval 0≤t≤2. g(t)=1+t for 0≤t≤2

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Chapter 5 Solutions

Find the average value of the function over the interval 0≤t≤2.

g(t)=1+t for 0≤t≤2