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Activity Test your understanding. Write TRUE on the space provided if the statement is correct and FALSE if incorrect. 1. y +1-2 dy, can be evaluated conveniently if we let y+1–2=t and dy = 4r(r +2)dt. dx = 2x-5 in (x+5)+c. 2. ydy 3. can be solved by either algebraic substitution or integration by parts. -1 V25 -x dx 4. In if we let u = /25 -x, then udu = -xdx. 5. In fx' /x? -4 ,we let y = Vx² – 4.

Activity Test your understanding. Write TRUE on the space provided if the statement is correct and FALSE if incorrect. 1. y +1-2 dy, can be evaluated conveniently if we let y+1–2=t and dy = 4r(r +2)dt. dx = 2x-5 in (x+5)+c. 2. ydy 3. can be solved by either algebraic substitution or integration by parts. -1 V25 -x dx 4. In if we let u = /25 -x, then udu = -xdx. 5. In fx' /x? -4 ,we let y = Vx² – 4.

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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Vx² +64dx sec' ede 6. 2 tan e 7. In x25 +x dx, we let x=5 tan e. S. In | x. -625 dx, if we let x= 25sece, thendx= 25sece de. 9. The result of the integral in 6, given 25 dx is 5 tan e- 50 + C. 10 (x/49-x² dx when converted to e is 7 sin e cos e de. 2 11. To each linear factor occurring once in the denominator of a proper fraction, Adx there corresponds a single partial fraction of the form where A is ax+b constant and to be determined. (4x +5x*)dx (x+1)(x2 -6x +34)2 12. By the method of partial fractions, the integral can be A B(2x - 6) evaluated when expressed as dx . x+1 (x2 - 6x + 34)2 (3x² +8x –12)kix x + 7x +12x 13. In the complete factored form of the denominator is x(x+ 4)(x+ 3)
Transcribed Image Text:Vx² +64dx sec' ede 6. 2 tan e 7. In x25 +x dx, we let x=5 tan e. S. In | x. -625 dx, if we let x= 25sece, thendx= 25sece de. 9. The result of the integral in 6, given 25 dx is 5 tan e- 50 + C. 10 (x/49-x² dx when converted to e is 7 sin e cos e de. 2 11. To each linear factor occurring once in the denominator of a proper fraction, Adx there corresponds a single partial fraction of the form where A is ax+b constant and to be determined. (4x +5x*)dx (x+1)(x2 -6x +34)2 12. By the method of partial fractions, the integral can be A B(2x - 6) evaluated when expressed as dx . x+1 (x2 - 6x + 34)2 (3x² +8x –12)kix x + 7x +12x 13. In the complete factored form of the denominator is x(x+ 4)(x+ 3)
Activity Test your understanding. Write TRUE on the space provided if the statement is correct and FALSE if incorrect. _1. y+1-2 dy,c can be evaluated conveniently if we let y+1-2 = t and dy = 4t(r +2)dt. dx = 2x- 5 In (x+5)+c. 2. ydy 3. can be solved by either algebraic substitution or integration by parts. 25-x dx 4. In if we let u = v25 – x, then udu = -xdx. 5. In [x/x² - 4 , we let y = Vx - 4.
Transcribed Image Text:Activity Test your understanding. Write TRUE on the space provided if the statement is correct and FALSE if incorrect. _1. y+1-2 dy,c can be evaluated conveniently if we let y+1-2 = t and dy = 4t(r +2)dt. dx = 2x- 5 In (x+5)+c. 2. ydy 3. can be solved by either algebraic substitution or integration by parts. 25-x dx 4. In if we let u = v25 – x, then udu = -xdx. 5. In [x/x² - 4 , we let y = Vx - 4.
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