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Evaluate 32x24 4x² + 6x + 5 First substitute u = Then 2 dx where 4x²+6x+50. 32x+24 dx 4x²+6x+5 Now integrate with respect to u to get 32x 24 So dx = 4x²+6x+5 + C du no + C

Evaluate 32x24 4x² + 6x + 5 First substitute u = Then 2 dx where 4x²+6x+50. 32x+24 dx 4x²+6x+5 Now integrate with respect to u to get 32x 24 So dx = 4x²+6x+5 + C du no + C

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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Evaluate 32x24 4x² + 6x + 5 First substitute u = Then 2 dx where 4x²+6x+50. 32x+24 dx 4x²+6x+5 Now integrate with respect to u to get 32x 24 So dx = 4x²+6x+5 + C du no + C
Transcribed Image Text:Evaluate 32x24 4x² + 6x + 5 First substitute u = Then 2 dx where 4x²+6x+50. 32x+24 dx 4x²+6x+5 Now integrate with respect to u to get 32x 24 So dx = 4x²+6x+5 + C du no + C
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