Problem

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

05:18

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 $

Answer

a. $\frac{5}{4}$
b. 1

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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.

Chris T.

Ohio State University

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Calculus: Early Transcendentals

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Problem

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

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 $

Answer

a. $\frac{5}{4}$
b. 1

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Video Transcript

Calculus: Early Transcendentals

Kayleah T.

Caleb E.

Samuel H.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

a. $\frac{5}{4}$b. 1

Calculus: Early Transcendentals

Kayleah T.

Caleb E.

Samuel H.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Kayleah T.

Caleb E.

Samuel H.

Joseph L.

a. $\frac{5}{4}$
b. 1