Explanation Given: ∫ 0 1 ln ( 4 − 2 x ) d x Key Concept Applied: Integration by parts is a method for evaluating a difficult integral. When the integral is a product of functions, the integration by parts formula moves the products out the equation so the integral can be solved more easily. ∫ u ⋅ d v = u v − ∫ v ⋅ d u 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: Integration by parts ∫ u ⋅ d v = u v − ∫ v ⋅ d u Formulas of Common Derivative ( c ) ' = 0 , where c any constant ( c ⋅ x ) ' = c ⋅ d x Formulas of Common Integral ∫ 1 u ⋅ d u = ln | u | + c ∫ ( f ( u ) ± g ( u ) ) ⋅ d u = ∫ f ( u ) ⋅ d u ± ∫ g ( u ) ⋅ d u ∫ a d x = a x ∫ c f ( u ) ⋅ d u = c ⋅ ∫ f ( u ) ⋅ d u Calculation: Description ∫ 0 1 ln ( 4 − 2 x ) d x Step 1: Set that inside function equal to u & v ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 8, Problem 21CRE

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To evaluate:

The given definite integral

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Chapter 8, Problem 20CRE

Chapter 8, Problem 22CRE

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Consider the graphs of y = f(x) and y = g(x) in the given diagram y= f(x). y = g(x) Evaluate (f+g)(2) -5 Determine all for which g(x) < f(x) Determine all for which f(x) +3 = g(x)

I) For what value(s) of x does g(x) = -4? Separate multiple answers with commas as needed. J) Give the interval(s) of such that g(x) > 0. Use the union symbol between multiple intervals. K) Give the interval(s) of such that g(x) <0. Use the union symbol between multiple intervals.

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Chapter 8 Solutions

CALCULUS (CLOTH)

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ 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

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The given definite integral

Solution Summary: The author explains the integration by parts formula for evaluating a difficult integral. U-Substitution is the reverse of the chain rule for differentiation.

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Chapter 8 Solutions

To evaluate:

The given definite integral