Explanation Given: The area between the x -axis and nonnegative function u ( x ) d v d x = h ( x ) is 15 on the interval [ 1 , 3 ] with u ( 3 ) = 0 . On the same interval the area between the x -axis and nonnegative function v ( x ) d u d x = k ( x ) is 20. Formula used: “If u and v are differentiable functions, then ∫ u d v = u v − ∫ v d u .” Calculation: Rearrange the formula and obtain, u v = ∫ v d u + ∫ u d v . Then, ∫ u ( x ) d v + ∫ v ( x ) d u = u ( x ) v ( x ) . The area between the x axis and the nonnegative function h ( x ) over the interval [ 1 , 3 ] is 15. That is ∫ 1 3 h ( x ) d x = 15 and ∫ 1 3 u ( x ) d v = 15 . The area between the x axis and the nonnegative function k ( x ) over the interval [ 1 , 3 ] is 20

Calculus with Applications (11th Edition)

11th Edition

ISBN: 9780321979421

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 8.1, Problem 34E

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The value of the function u(x)v(x) at x=1 .

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Chapter 8.1, Problem 33E

Chapter 8.1, Problem 35E

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Chapter 8 Solutions

Calculus with Applications (11th Edition)

Ch. 8.1 - YOUR TURN 1 Find Ch. 8.1 - YOUR TURN 2 Find Ch. 8.1 - YOUR TURN 3 Find Ch. 8.1 - YOUR TURN 4 Find . Ch. 8.1 - YOUR TURN 5 Find Ch. 8.1 - Prob. 1WE Ch. 8.1 - Find the following. W2. Ch. 8.1 - Find the following. W3. Ch. 8.1 - Find the following. W4. Ch. 8.1 - Use integration by parts to find the integrals in...

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The value of the function u ( x ) v ( x ) at x = 1 .

The value of the function u(x)v(x) at x=1 .

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Chapter 8 Solutions