_1. y +1-2 dy, can be evaluated conveniently if we let y+1-2=t and dy = 4r(r +2)dt. %3!

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...
icon
Related questions
Question
100%

Answer it with solution. Thank you. 

Vx? + 64dx
6.
sec ede
2
tan ? e
7. In x/25 + x* dx, we let x=5 tan e.
8. In [x Jx? -625 dx, if we let x= 25sece, thendx=25seco de.
9. The result of the integral in 6 , given
25
dx is 5 tan e- 50+ C.
_10. (x/49-x²dx_when converted to e is75 sin 2 e cos e de.
_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.
(4x2 +5x*)dx
_12. By the method of partial fractions, the integral
can be
(x+1)(x2 -6x +34)2
A
В(2х - 6)
evaluated when expressed as
dx.
x+1
(x2 - 6x + 34)2
(3x + 8x –12 ax
x' + 7x +12x
13. In
the complete factored form of the denominator is
x(x+ 4)(x+3)
Transcribed Image Text:Vx? + 64dx 6. sec ede 2 tan ? e 7. In x/25 + x* dx, we let x=5 tan e. 8. In [x Jx? -625 dx, if we let x= 25sece, thendx=25seco de. 9. The result of the integral in 6 , given 25 dx is 5 tan e- 50+ C. _10. (x/49-x²dx_when converted to e is75 sin 2 e cos e de. _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. (4x2 +5x*)dx _12. By the method of partial fractions, the integral can be (x+1)(x2 -6x +34)2 A В(2х - 6) evaluated when expressed as dx. x+1 (x2 - 6x + 34)2 (3x + 8x –12 ax 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 , can be evaluated conveniently if we let y+1-2=t and
dy = 4r( +2)dt.
dx
= 2x-5n (+5)+c.
2.
Vx+5
ydy
can be solved by either algebraic substitution or integration by parts.
3.
V25 -x* dx
4. In
if we let u = v25 – x, then udu = -xdx.
5. In [x/x? - 4 , we lety = Jx? - 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 , can be evaluated conveniently if we let y+1-2=t and dy = 4r( +2)dt. dx = 2x-5n (+5)+c. 2. Vx+5 ydy can be solved by either algebraic substitution or integration by parts. 3. V25 -x* dx 4. In if we let u = v25 – x, then udu = -xdx. 5. In [x/x? - 4 , we lety = Jx? - 4.
Expert Solution
steps

Step by step

Solved in 5 steps with 5 images

Blurred answer
Knowledge Booster
Linear Equations
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning