Explanation Given: ∫ x 2 d x ( x + 5 ) 3 Key Concept Applied: U-Substitution is a technique we use when the integrand is a composite function. A composite function is a function inside another function of the form f ( g ( x ) ) where f ( x ) is the outside function and g ( x ) is the inside function. The first thing we will need to do is to recognize that we are being asked to integrate a product of a function and its derivative, and it takes the form of a composite function. This can be generalized for any product that is a composite function multiplied by the derivative of the inner function. This integration rule follows from the chain rule for differentiation, because U-Substitution is nothing more than the reverse of the chain rule. Substitution Rule To integrate: ∫ f ( g ( x ) ) ⋅ g ' ( x ) ⋅ d x ↗ ↖ I n s i d e f u n c t i o n O u t s i d e F u n c t i o n ( D e r i v a t i v e o f I n s i d e ) The essence of the substitution rule is to take an integral in terms of x and transform or change it into terms of u by finding the outside function and inside function, where the outside function is the derivative of the inside function. Then we will make a suitable substitution that will simplify our integrand so that we can integrate, as illustrated in three easy steps below: ∫ f ( g ( x ) ) ⋅ g ' ( x ) ⋅ d x , I f u = g ( x ) ⇒ d u = g ' ( x ) ⋅ d x , T h e n d x = d u g ' ( x ) A f t e r U − S u b t i t u t i o n , I n t e g r a l w i l l b e c o m e ... ∫ f ( u ) ⋅ g ' ( x ) ⋅ d u g ' ( x ) C a n c e l l i n g g ' ( x ) , w e g e t ... ∫ f ( u ) ⋅ d u Formula used: Exponent Rules Rule 1 Rule 2 Rule 3 Rule 4 a = a 1 2 a 3 = a 1 3 a m n = a m n 1 a m = a − m Rules used here in this problem are highlighted in green in the below table: Basic Properties/Rules/Formulas of Common Derivative ( f ( x ) ) ' notation indicates derivative of function Polynomials Function ( c ) ' = 0 , where c any constant ( x ) ' = d x ( c ⋅ x ) ' = c ⋅ d x ( x n ) ' = n ⋅ x n − 1 ⋅ d x ( c ⋅ x n ) ' = c ⋅ n ⋅ x n − 1 ⋅ d x Trigonometric Functions ( sin x ) ' = cos x ⋅ d x ( cos x ) ' = − sin x ⋅ d x ( tan x ) ' = sec 2 x ⋅ d x ( sec x ) ' = sec x ⋅ tan x ⋅ d x ( csc x ) ' = − csc x ⋅ cot x ⋅ d x ( cot x ) ' = − csc 2 x ⋅ d x Inverse Trigonometric Functions ( sin − 1 x ) ' = 1 1 − x 2 ⋅ d x ( cos − 1 x ) ' = − 1 1 − x 2 ⋅ d x ( tan − 1 x ) ' = 1 1 + x 2 ⋅ d x ( sec − 1 x ) ' = 1 | x | ⋅ x 2 − 1 ⋅ d x ( csc − 1 x ) ' = − 1 | x | ⋅ x 2 − 1 ⋅ d x ( cot − 1 x ) ' = − 1 1 + x 2 ⋅ d x Exponential Functions ( a x ) ' = a x ⋅ ln a ⋅ d x ( e x ) ' = e x ⋅ d x Logarithmic Functions ( ln ( x ) ) ' = 1 x ⋅ d x , x > 0 ( ln | x | ) ' = 1 x ⋅ d x , x ≠ 0 ( log a ( x ) ) ' = 1 x ln a ⋅ d x , x > 0 Basic Properties/Rules/Formulas of Common Integral Polynomials Function ∫ c f ( u ) ⋅ d u = c ⋅ ∫ f ( u ) ⋅ d u ∫ ( f ( u ) ± g ( u ) ) ⋅ d u = ∫ f ( u

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 5.7, Problem 44E

To determine

To evaluate:

The given indefinite integral

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Chapter 5.7, Problem 43E

Chapter 5.7, Problem 45E

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Producers' Surplus A supplier of portable hair dryers will make x hundred units of hair dryers available in the market when the unit price is P169+ 4x dollars. Determine the producers' surplus if the market price is set at $17/unit. (Round your answer to two decimal places.) $50547.33

Consumers' Surplus The demand function for a certain make of replacement cartridges for a water purifier is given by the following equation where p is the unit price in dollars and x is the quantity demanded each week, measured in units of a thousand. p = -0.01x2 -0.1x + 19 Determine the consumers' surplus if the market price is set at $7/cartridge. (Round your answer to two decimal places.) $2250

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Chapter 5 Solutions

CALCULUS (CLOTH)

Ch. 5.1 - Prob. 1PQ Ch. 5.1 - Prob. 2PQ Ch. 5.1 - Prob. 3PQ Ch. 5.1 - Prob. 4PQ Ch. 5.1 - Prob. 5PQ Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E

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Prob. 31E Ch. 5.7 - Prob. 32E Ch. 5.7 - Prob. 33E Ch. 5.7 - Prob. 34E Ch. 5.7 - Prob. 35E Ch. 5.7 - Prob. 36E Ch. 5.7 - Prob. 37E Ch. 5.7 - Prob. 38E Ch. 5.7 - Prob. 39E Ch. 5.7 - Prob. 40E Ch. 5.7 - Prob. 41E Ch. 5.7 - Prob. 42E Ch. 5.7 - Prob. 43E Ch. 5.7 - Prob. 44E Ch. 5.7 - Prob. 45E Ch. 5.7 - Prob. 46E Ch. 5.7 - Prob. 47E Ch. 5.7 - Prob. 48E Ch. 5.7 - Prob. 49E Ch. 5.7 - Prob. 50E Ch. 5.7 - Prob. 51E Ch. 5.7 - Prob. 52E Ch. 5.7 - Prob. 53E Ch. 5.7 - Prob. 54E Ch. 5.7 - Prob. 55E Ch. 5.7 - Prob. 56E Ch. 5.7 - Prob. 57E Ch. 5.7 - Prob. 58E Ch. 5.7 - Prob. 59E Ch. 5.7 - Prob. 60E Ch. 5.7 - Prob. 61E Ch. 5.7 - Prob. 62E Ch. 5.7 - Prob. 63E Ch. 5.7 - Prob. 64E Ch. 5.7 - Prob. 65E Ch. 5.7 - Prob. 66E Ch. 5.7 - Prob. 67E Ch. 5.7 - Prob. 68E Ch. 5.7 - Prob. 69E Ch. 5.7 - Prob. 70E Ch. 5.7 - Prob. 71E Ch. 5.7 - Prob. 72E Ch. 5.7 - Prob. 73E Ch. 5.7 - Prob. 74E Ch. 5.7 - Prob. 75E Ch. 5.7 - Prob. 76E Ch. 5.7 - Prob. 77E Ch. 5.7 - Prob. 78E Ch. 5.7 - Prob. 79E Ch. 5.7 - Prob. 80E Ch. 5.7 - 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The given indefinite integral

Solution Summary: The author explains how U-Substitution is used when the integrand is a composite function. The essence of the substitution rule is to take an integral and transform it into terms of u.

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To evaluate: The given indefinite integral

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Chapter 5 Solutions

To evaluate:

The given indefinite integral