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PROBLEMS In Problems 1–52, fir 1. 7 dx x° dx 5x- dx 7. dx x7 3. 5.

PROBLEMS In Problems 1–52, fir 1. 7 dx x° dx 5x- dx 7. dx x7 3. 5.

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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3 and 5 

Solution: does not fit a integra Solution: can break up the by term in the CAUTIONA In Example 8, we first multiplied the factors in the integrand. The answer could not have been found simply in terms of f y dy and f (y+) dy. There is not a formula for the integral of a general product of functions. plying the integrand we get multi- 25. 2' 2y3 +C 27. +C = 4 4. 6 29. 31. Using Algebraic Manipulation to Find an Indefinite Integral EXAMPLE 9 33 / (2r - 1)(x+3) *xp 3. Solution: By factoring out the constant and multiplying the binomials, we got (2x-1)(x+3) -| = xp (2x2 + 5x - 3) dx 9. 1. + (5) (2) 3 + C 9. 5x2 - 2 ) Another algebraic approach to part (b) is 12 2. - 1 dx. -2 xp b. Find maini gaiinO Solution: We can break up the integrand into fractions by dividing each term in a numerator by the denominator: and so on. xp(x-x) 1. dx (x – x-2) dx bris mus s to = xp =ɔ+ 2. 1. %3D ɔ+- + Now Work Problem 49 PROBLEMS 14.2 In Problems 1–52, find the indefinite integrals. Finding 2. 1. t7/4 1. 7 dx 9. 10. xp 2x9/4 xp - 3. 4. 5x24 dx 11. (4+ t) dt (7r5 + 4r2 + 1) dr 12. 5. -7 6. zp 3 - 5y) dy &p (^s – 13. 14. |(5 – 2w – 6w²) dw 7. dx 8. 15. | (312 - 4t + 5) dt 16. xP X | IP („7 + µJ + zJ + I) |
Transcribed Image Text:Solution: does not fit a integra Solution: can break up the by term in the CAUTIONA In Example 8, we first multiplied the factors in the integrand. The answer could not have been found simply in terms of f y dy and f (y+) dy. There is not a formula for the integral of a general product of functions. plying the integrand we get multi- 25. 2' 2y3 +C 27. +C = 4 4. 6 29. 31. Using Algebraic Manipulation to Find an Indefinite Integral EXAMPLE 9 33 / (2r - 1)(x+3) *xp 3. Solution: By factoring out the constant and multiplying the binomials, we got (2x-1)(x+3) -| = xp (2x2 + 5x - 3) dx 9. 1. + (5) (2) 3 + C 9. 5x2 - 2 ) Another algebraic approach to part (b) is 12 2. - 1 dx. -2 xp b. Find maini gaiinO Solution: We can break up the integrand into fractions by dividing each term in a numerator by the denominator: and so on. xp(x-x) 1. dx (x – x-2) dx bris mus s to = xp =ɔ+ 2. 1. %3D ɔ+- + Now Work Problem 49 PROBLEMS 14.2 In Problems 1–52, find the indefinite integrals. Finding 2. 1. t7/4 1. 7 dx 9. 10. xp 2x9/4 xp - 3. 4. 5x24 dx 11. (4+ t) dt (7r5 + 4r2 + 1) dr 12. 5. -7 6. zp 3 - 5y) dy &p (^s – 13. 14. |(5 – 2w – 6w²) dw 7. dx 8. 15. | (312 - 4t + 5) dt 16. xP X | IP („7 + µJ + zJ + I) |
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