Explanation Given: ∫ 4 x − 1 ⋅ d x , u = 4 x − 1 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 −

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 5.7, Problem 14E

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The given integral in terms of $u$ and $d u$ , and then evaluate it.

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Chapter 5.7, Problem 13E

Chapter 5.7, Problem 15E

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Let the region R be the area enclosed by the function f(x)= = 3x² and g(x) = 4x. If the region R is the base of a solid such that each cross section perpendicular to the x-axis is an isosceles right triangle with a leg in the region R, find the volume of the solid. You may use a calculator and round to the nearest thousandth. y 11 10 9 00 8 7 9 5 4 3 2 1 -1 -1 x 1 2

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Chapter 5 Solutions

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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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The given integral in terms of u and d u , and then evaluate it.

Solution Summary: The author explains that U-Substitution is a technique used to evaluate an integral in terms of u and d.

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The given integral in terms of $u$ and $d u$ , and then evaluate it.

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Chapter 5 Solutions

The given integral in terms of u and d u , and then evaluate it.

Solution Summary: The author explains that U-Substitution is a technique used to evaluate an integral in terms of u and d.

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To Write: The given integral in terms of u and du, and then evaluate it.

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Chapter 5 Solutions

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The given integral in terms of $u$ and $d u$ , and then evaluate it.