Problem

Show that $$ 0 \le \int^{10}_5 \frac{x^2}{x^4 + x…

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Problem 80 Hard Difficulty

(a) Show that $ \cos (x^2) \ge \cos x $ for $ 0 \le x \le 1 $.
(b) Deduce that $ \displaystyle \int^{\pi/6}_0 \cos (x^2) \, dx \ge \frac{1}{2} $.

Answer

a.$x \in[0,1], \quad x^{2} \leq x$
b.$\frac{1}{2}$

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Video Transcript

in part A were asked to show that the coastline of X squared is greater than or equal to cosine of X for X between zero and one you know, he was boiling some guys. Well first listen, we gotta we gotta get that water to after all observe at the function F. Of X which is equal to the coastline of X is decreasing. They for sure on the interval 0 to Pi over two. This is easy to see graphically. Alternatively, you could prove this by taking the derivative F. Prime of X, which is equal to negative sine of X. You never know who did which of course this is going to be less than zero on the interval zero to pi over too. Well, the open intervals europe I refer to he said no And therefore it also implies that co sign of X is decreasing on the interval inside this 01. We also know that X squared is less than X mm On the interval 01 I guess I should say the open interval technically. So this is different from when X is greater than one, where X squared is always greater than X. Yeah, teaching a feared and respected. Now if we apply this well, because F cosine of X was decreasing, it follows that the co sign of X squared is therefore going to be greater than the co sign of X. Since X squared is less than X, it is a voice be drawn. Now, this is only on the open interval 01 you're saying in particular, we have that co sign of zero squared is the same as the co sign of zero and the co sign of one squared equals the co sign of one and therefore co sign of X squared Is greater than or equal to the co sign of X. on the closed interval 01. This is what we wanted to prove in part A then in part B were asked to deduce that the integral from zero to pi over six of cosine of X squared Is greater than or equal to 1/2. That's what like direct. Well, we'll use the comparison theorem for integral. So of course, Hi over six lies between zero and 1 and therefore it follows that the co sign of X squared is greater than or equal to the coast sine of X on the closed interval zero to pi over six. But I don't think the genie should. So we have the integral from zero to pi over six of the co sign of X squared dx is going to be great and equal to the integral from zero to pi over six of just the co sign of X dx mm. Now the left hand side, we don't know a simple anti derivative for co sign of X squared. We do know we can rewrite the right hand side As the integral from 0-6. While we take the anti derivative, this is now sine of X, evaluated from zero to pi over six, which is the sine of pi over six, which is one half minus the sine of zero, which is zero is simply one half and therefore simplifying the integral from zero to pi over six of the co sign of X squared dx Must be greater than or equal to 1/2. So in this way we obtain an estimate for an integral, which it is not easy to find directly. I want to a fucking diarrhea.

Chris T.

Ohio State University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Problem

Show that $$ 0 \le \int^{10}_5 \frac{x^2}{x^4 + x…

Problem 80 Hard Difficulty

(a) Show that $ \cos (x^2) \ge \cos x $ for $ 0 \le x \le 1 $.
(b) Deduce that $ \displaystyle \int^{\pi/6}_0 \cos (x^2) \, dx \ge \frac{1}{2} $.

Answer

a.$x \in[0,1], \quad x^{2} \leq x$
b.$\frac{1}{2}$

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Calculus: Early Transcendentals

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Video Transcript

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

a.$x \in[0,1], \quad x^{2} \leq x$b.$\frac{1}{2}$

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

Catherine R.

Anna Marie V.

Kristen K.

Samuel H.

a.$x \in[0,1], \quad x^{2} \leq x$
b.$\frac{1}{2}$