Explanation Given: The function r and s are everywhere differentiable and that u ( 3 ) = 0 .And 1 ≤ x ≤ 3 the area between the x -axis , and the nonnegative function h ( x ) = u ( x ) d v d x is 15, and that on the same interval the area between the x -axis and the nonnegative function k ( x ) = v ( x ) d u d x is 20. Approach: The area between curve y = f ( x ) and the x - axis from x = a to x = b : A = ∫ a b f ( x ) d x The integration of by parts ∫ u d v = u v − ∫ v d u Here u ( x ) and v ( x ) differential function. Calculation: Consider the nonnegative function, h ( x ) = u ( x ) d v ( x ) d x Integrate the given function from x = 1 to x = 3 that yield area: ∫ 1 3 h ( x ) d x = ∫ 1 3 u ( x ) d v ( x ) d x d x Substitute 15 for ∫ 1 3 h ( x ) d x in the above equation, ∫ 1 3 u ( x ) d v d x d x = 15 Apply the integration by part,

Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

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ISBN: 9780137553457

Author: Raymond Greenwell, Nathan Ritchey

Publisher: PEARSON+

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Chapter 8.2, Problem 46E

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The value of function u(1)v(1) such that the function r and s are everywhere differentiable.

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Chapter 8.2, Problem 45E

Chapter 8.2, Problem 47E

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Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

Ch. 8.1 - YOUR TURN Use the trapezoidal rule with n=4 to...Ch. 8.1 - Prob. 2YT Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E

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The value of function u ( 1 ) v ( 1 ) such that the function r and s are everywhere differentiable.

Solution Summary: The author explains that the value of function u(1)v (1,1) is -35. The area between the x-axis and the nonnegative function is

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To find:

The value of function u(1)v(1) such that the function r and s are everywhere differentiable.

Want to see the full answer?

Chapter 8 Solutions

The value of function u ( 1 ) v ( 1 ) such that the function r and s are everywhere differentiable.

Solution Summary: The author explains that the value of function u(1)v (1,1) is -35. The area between the x-axis and the nonnegative function is

Videos

To find: The value of function u(1)v(1) such that the function r and s are everywhere differentiable.

Want to see the full answer?

Chapter 8 Solutions

To find:

The value of function u(1)v(1) such that the function r and s are everywhere differentiable.