Explanation Given information: The function f ( t ) = 1000 + 200 sin 2 ( t − π 4 ) where 0 ≤ t ≤ 3 π 4 Formula used: The average of the function f ( t ) where a ≤ t ≤ b is 1 b − a ( ∫ a b f ( t ) d t ) . ∫ cos ( a x + b ) d x = 1 a sin ( a x + b ) + C , where C is the constant of integration. Calculation: The average of the function f ( t ) = 1000 + 200 sin 2 ( t − π 4 ) where 0 ≤ t ≤ π is 1 3 π 4 − 0 ( ∫ 0 3 π / 4 ( 1000 + 200 sin 2 ( t − π 4 ) ) d t ) . = 1 3 π / 4 ( ∫ 0 3 π / 4 1000 d t + ∫ 0 3 π / 4 200 sin 2 ( t − π 4 ) d t ) = 1 3 π / 4 ( 1000 ∫ 0 3 π / 4 1 d t + 200

Pearson eText for Calculus & Its Applications -- Instant Access (Pearson+)

15th Edition

ISBN: 9780137590728

Author: Larry Goldstein, David Lay

Publisher: PEARSON+

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Chapter 8, Problem 73RE

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To calculate: Theaverage of the function f(t)=1000+200 sin 2(t−π4) where 0≤t≤π.

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Chapter 8, Problem 72RE

Chapter 8, Problem 74RE

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Theaverage of the function f ( t ) = 1000 + 200 sin 2 ( t − π 4 ) where 0 ≤ t ≤ π .

Solution Summary: The author calculates the average of the function f(t)=1000+200sin2( t-pi 4).

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To calculate: Theaverage of the function f(t)=1000+200 sin 2(t−π4) where 0≤t≤π.

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Chapter 8 Solutions