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(d) | sin(ax + b) dx = –- 1 cos(a.x + b) + C %3D a (e) / (ax+b)" dx 1 (ax + b)"+1 + C а (п + 1)

(d) | sin(ax + b) dx = –- 1 cos(a.x + b) + C %3D a (e) / (ax+b)" dx 1 (ax + b)"+1 + C а (п + 1)

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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solve (d e)

Question 4 Verify the following antidifferentiation formulas. (a) / (b) / In x dx = x ln x – x + C tan x dx = - In | cos a| + C COS 1 ar+b dx = ear+b + C a (d) / sin(ax + b) dæ 1 cos(ax + b) + C a (e) / 1 (a.x + b)"+1 + C (ax + b)" d.x = а(n + 1) The last three are cases of “baby substitution" the chain rule with a linear inside inside function ax + b establishes 1 | f'(a.x + b) da = =f(a.x + b) + C. d (ax+b) dx do not do this with more complicated This idea works because = a is a constant function inside functions! er? is not an antiderivative of e*- 2x For example, the quotient rule says that 2r (e*2x) d d.x 2.2 t ez? 2x 4x2 In fact, it is known that we will never obtain a “nice formula" for dx in the same way we can for 1 e2" + C. e2x d.x = 2
Transcribed Image Text:Question 4 Verify the following antidifferentiation formulas. (a) / (b) / In x dx = x ln x – x + C tan x dx = - In | cos a| + C COS 1 ar+b dx = ear+b + C a (d) / sin(ax + b) dæ 1 cos(ax + b) + C a (e) / 1 (a.x + b)"+1 + C (ax + b)" d.x = а(n + 1) The last three are cases of “baby substitution" the chain rule with a linear inside inside function ax + b establishes 1 | f'(a.x + b) da = =f(a.x + b) + C. d (ax+b) dx do not do this with more complicated This idea works because = a is a constant function inside functions! er? is not an antiderivative of e*- 2x For example, the quotient rule says that 2r (e*2x) d d.x 2.2 t ez? 2x 4x2 In fact, it is known that we will never obtain a “nice formula" for dx in the same way we can for 1 e2" + C. e2x d.x = 2
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