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Using the method of substitution, I = [ x²(1 + 2x³¹)³dx = [ f(u)du where u= du = f(u) = Using the above information: [ x²(1 + 2x³)³dx - f(u)du = || da (in terms (in terms

Using the method of substitution, I = [ x²(1 + 2x³¹)³dx = [ f(u)du where u= du = f(u) = Using the above information: [ x²(1 + 2x³)³dx - f(u)du = || da (in terms (in terms

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Using the method of substitution, I ! = [ z²(1 + 2x³)³dx = [ f(u)du where U= du: = f(u) = Using the above information: [ x²(1 + 2x³)²dx = [ f(u)du = || dx (in terms of u) (in terms of x) Note: Don't forget the constant of integration in the last two blanks.
Transcribed Image Text:Using the method of substitution, I ! = [ z²(1 + 2x³)³dx = [ f(u)du where U= du: = f(u) = Using the above information: [ x²(1 + 2x³)²dx = [ f(u)du = || dx (in terms of u) (in terms of x) Note: Don't forget the constant of integration in the last two blanks.
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u= 1+2x^3 -lu = 6x^2 f(u) = 4^2/6 Using the above information: [ x² (1+2x³)²dx = [ f(u)du 4^3/10 + C (1+2x^3)^3/10 +C dx (in terms of u) (in terms of x)
Transcribed Image Text:u= 1+2x^3 -lu = 6x^2 f(u) = 4^2/6 Using the above information: [ x² (1+2x³)²dx = [ f(u)du 4^3/10 + C (1+2x^3)^3/10 +C dx (in terms of u) (in terms of x)
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