Question

Answered step-by-step

Problem 82 Hard Difficulty

Show that if $ a > -1 $ and $ b > a + 1 $, then the following integral is convergent
$$ \int_0^\infty \frac{x^a}{1 + x^b}\ dx $$

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Wen Zheng

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 8

Improper Integrals

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 7

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Video Transcript

The problem is show that if its greatest on ninety one and B is created on a past one, then the following into grow is converted. This improper integral we can, right? It's integral. Is he called to integral from zero to one ax to a over one plus X to be the eggs class into girl from one to the infinity of ax to a for one flat sachs to be power? Yes, on the first part is is a definite integral, so it will be a constant a number. So if the second heart is converted, then are integral from zero to infinity of dysfunction is commitment. Now look at this improper integral. We can write disfunction, eyes divided acts to its power for the numerator and the denominator. So this is one over one. Over. Ax to ace Power class Act two I mean minus a Yes. This function is always smaller, Czar, into your from one to infinity. One over x So e minus its power is it for this function? Denominator won over acts to ace power plus acts to humanise, eh is greeted on axe to be minus his power. These axes, critters and zero. Now look at this improper anti girl Face B is created on a past one. It's a B minus one. It's three to one. So this improper anti grow is commitment. And by the comparison, cereal thiss improper Anti Girl is also important is an integral from zero to infinity of the function axed weigh over one plus x two b's power is also converted.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Joseph Lentino

Calculus 2 / BC Courses

Recommended Videos

Watch More Solved Questions in Chapter 7

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Joseph Lentino

James Stewart

Calculus: Early Transcendentals