Explanation Given: The given integral is ∫ − 2 4 f ( x ) d x and ∫ − 2 4 g ( x ) d x . Moreover,the following results are given: ∫ − 2 4 [ 5 f ( x ) + 2 g ( x ) ] d x = 7 ∫ − 2 4 [ 3 f ( x ) + g ( x ) ] d x = 4 Concept used: If f and g are integrable on [ a , b ] , then c f ( x ) ± d g ( x ) is also integrable for any constant c and d . Linear property of definite integral are as follows: ∫ a b [ c f ( x ) ± d g ( x ) ] d x = c ∫ a b f ( x ) d x ± d ∫ a b g ( x ) d x Formula used: Sub-division Property of definite integral: ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x Where c ∈ [ a , b ] Calculation: Consider the following: ∫ − 2 4 [ 5 f ( x ) + 2 g ( x ) ] d x = 7 By linearity property, ∫ − 2 4 [ 5 f ( x ) + 2 g ( x ) ] d x = 7 5 ∫ − 2 4 f ( x ) d x + 2 ∫ − 2 4 g ( x ) d x = 7 ............. (1) Again, consider the following: ∫ − 2 4 [ 3 f ( x ) + g ( x ) ] d x = 4 By linearity property, ∫ − 2 4 [ 3 f ( x ) + g ( x ) ] d x = 4 3 ∫ − 2 4 f ( x ) d x + ∫ − 2 4 g ( x ) d x = 4

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 5.3, Problem 43PS

To determine

To find: The value of the integral ∫−24f(x)dx and ∫−24g(x)dx .

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Chapter 5.3, Problem 42PS

Chapter 5.3, Problem 44PS

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Chapter 5 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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The value of the integral ∫ − 2 4 f ( x ) d x and ∫ − 2 4 g ( x ) d x .

Solution Summary: The author explains that the value of the integral is displaystyle, int, and_-24gleft, respectively.

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To find: The value of the integral ∫−24f(x)dx and ∫−24g(x)dx .

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Chapter 5 Solutions