Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Sent to:
Search glass icon
  • Login
  • Textbooks
  • Ask our Educators
  • Study Tools
    Study Groups Bootcamps Quizzes AI Tutor iOS Student App Android Student App StudyParty
  • For Educators
    Become an educator Educator app for iPad Our educators
  • For Schools

Problem

(a) Show that $ \cos (x^2) \ge \cos x $ for $ 0 \…

05:18

Question

Answered step-by-step

Problem 79 Hard Difficulty

(a) Show that $ 1 \le \sqrt{1 + x^3} \le 1 + x^3 $ for $ x \ge 0 $.
(b) Show that $ \displaystyle 1 \le \int^1_0 \sqrt{1 + x^3} \, dx \le 1.25 $


Video Answer

Solved by verified expert

preview
Numerade Logo

This problem has been solved!

Try Numerade free for 7 days

Chris Trentman
Numerade Educator

Like

Report

Textbook Answer

Official textbook answer

Video by Chris Trentman

Numerade Educator

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

More Answers

01:17

Frank Lin

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Related Topics

Integrals

Integration

Discussion

You must be signed in to discuss.
Top Calculus 1 / AB Educators
Grace He
Anna Marie Vagnozzi

Campbell University

Kayleah Tsai

Harvey Mudd College

Michael Jacobsen

Idaho State University

Calculus 1 / AB Courses

Lectures

Video Thumbnail

05:53

Integrals - Intro

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

Video Thumbnail

40:35

Area Under Curves - Overview

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

Join Course
Recommended Videos

05:15

\begin{equation}
\begin…

01:02

(a) Show that $1 \leqslant…

03:45

Show that $1 \leq \int_{0}…

01:22

Show that

01:22

Show that

Watch More Solved Questions in Chapter 5

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
Problem 83
Problem 84
Problem 85
Problem 86

Video Transcript

zanies Nashville 20. We are asked to show that the inequalities one is less than or equal to the square root of one plus X. Cute. Less than or equal to one plus X. Cute is true. This is part a prove this inequality first. We want to show the left inequality that one is less than or equal to the square root of one plus X cubed for non negative X. Well, clearly one is less than or equal to one plus X cubed. This is because X cubed is greater than or equal to zero. Mr in this case mm And therefore it follows that one Which is the same as the square root of one is less than equal to the square root of one plus x cubed. At least for X cube X greater than or equal to zero. Especially exact problem now lot of guys. Yeah. Actually there's no shame in that. No one who listens all right. Since one plus X cubed is greater than or equal to one. It follows that the square to one for six cubed is greater than or equal to one two. But I'll call I think I have turf toe now. Yeah, I think it's from church. You we know that if a number is greater than or equal to one and that number is going to be less than equal to its square. Therefore it follows that the square root of one plus x cubed Because one plus execute is great and the little one. Well, this has to be less than or equal to the square root of one plus X cubed squared. This is the same as one plus X. Cute. So combining these statements, it follows that for X greater than or equal to zero. One is less than or equal to the square root of one plus x cubed which is less than or equal to one plus x. cute. We got mm hmm. Asian guy check it then in part view. Uh He's gone. Excuse me. Can I see your I. D. Sir? He hasn't been there. My favorite part is can I see your I. D. Please sir? And I'm like I'm on the show were asked to prove the inequality that one is less than or equal to the integral from 0 to 1 of the square root of one plus X cubed dx which is less than or equal to 1.25 So we're going to use the inequality from Part A in particular will use the comparison property of definite integral. So from Part A we have that one is less than or equal to the square root of one plus X cubed. Which is just equal to one plus X cube. This was valid for all non negative X. But it's also true in particular for X between zero and one. There you go. We should cost play for that played. Uh and therefore by a comparison property we have that. The integral from 0 to 1 of one dx is going to be less than or equal to the integral from 0 to 1 of the square root of one plus x cubed dx which is less than equal to the integral from 0 to 1 of one plus X cube dx. And evaluating these integral wells. Well, this is taking anti derivatives X from 0 to 1 is less than equal to the integral from 0 to 1 of the square root of one plus x cubed dx less than or equal to x plus 1/4 Extra forth from 0 to 1. And now evaluating these expressions, we get that one is less than or equal to the integral from 0 to 1 of the square root of one plus X cubed dx which is less than or equal to. And this is one plus 1/4 or one plus 10.25 which is 1.25 So we get the inequality one is less than or equal to the integral from 0 to 1 of the square root of one plus X cubed dx, which is less than or equal to 1.25 In this way we get an approximation for this integral for which it is not easy to find the exact value.

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.
Math (Geometry, Algebra I and II) with Nancy
Arrow icon
Participants icon
178
Hosted by: Ay?Enur Çal???R
Math (Algebra 2 & AP Calculus AB) with Yovanny
Arrow icon
Participants icon
75
Hosted by: Alonso M
See More

Related Topics

Integrals

Integration

Top Calculus 1 / AB Educators
Grace He

Numerade Educator

Anna Marie Vagnozzi

Campbell University

Kayleah Tsai

Harvey Mudd College

Michael Jacobsen

Idaho State University

Calculus 1 / AB Courses

Lectures

Video Thumbnail

05:53

Integrals - Intro

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

Video Thumbnail

40:35

Area Under Curves - Overview

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

Join Course
Recommended Videos

05:15

\begin{equation} \begin{array}{l}{\text { (a) Show that } 1 \leqslant \sqrt{1+…

01:02

(a) Show that $1 \leqslant \sqrt{1+x^{3}} \leqslant 1+x^{3}$ for $x \geqslant 0…

03:45

Show that $1 \leq \int_{0}^{1} \sqrt{1+x^{4}} d x \leq \frac{6}{5} .$ Hint: Exp…

01:22

Show that

01:22

Show that
Additional Mathematics Questions

01:37

'please quickly solve this problem
X
Ji ~7+*5 %z
Hen fne 9+ 4&…

05:27

'Show that multiplication of matrices is not commutative by determining…

01:05

'a wooden belan was made from a big cylinder and 2 equal small cylinde…

00:57

'Find the angle measure X in the following figure.
60o
80o 1209

00:21

'Type the correct answer in the box. In the above triangle, cosA/cosB =…

01:19

"hey Please solve very very urgent..Jo solve karega me follow karungi o…

02:13

'Let A be the set of all even natural numbers (including zero), B be th…

01:24

'Express each sum or difference as a single logarithm'

00:55

'a. What is the theoretical probability of rolling a number greater tha…

00:55

'a. What is the theoretical probability of rolling a number greater tha…

Add To Playlist

Hmmm, doesn't seem like you have any playlists. Please add your first playlist.

Create a New Playlist

`

Share Question

Copy Link

OR

Enter Friends' Emails

Report Question

Get 24/7 study help with our app

 

Available on iOS and Android

About
  • Our Story
  • Careers
  • Our Educators
  • Numerade Blog
Browse
  • Bootcamps
  • Books
  • Notes & Exams NEW
  • Topics
  • Test Prep
  • Ask Directory
  • Online Tutors
  • Tutors Near Me
Support
  • Help
  • Privacy Policy
  • Terms of Service
Get started