Using properties of Integrals: show 1/96 <= (1/pi^2)....

[ku1005]

I've been working on this one for a while. I can see where the numbers come from, but I just can't seem to justify why it would be this way...? Any tips/suggestions woul be great!

. . .Show that:

. . .1/96 <= 1/(pi^2) Int(Pi/4 to Pi/3) xcosx dx <= 1/36sqrt(2)

I have been trying to use the Squeeze Theorem, so:

. . .-x/pi^2 <= xcosx/pi^2 <= x/pi^2

Thus, when x is between pi/4 and pi/3, we get:

. . .1/4pi <= xcosx/pi^2 <= 1/3pi

Therefore:

. . .1/48 <= Int (pi/4 to Pi/3) xcosx/pi^2 <= 1/36 (ie I multiplied by pi/12)

I am out by a factor of 1/2 (ie i understand is cos(pi/3) on the left and 1/sqrt(2) - ie cos(pi/4)on the right but dont understand how to encorporate them???

Cheers for any suggestions and tips!

Rhys

 

[pka]

It may help to note that on the interval is question cosine is decreasing.
\(\displaystyle \L\frac{\pi }{4} \le x \le \frac{\pi }{3}\, \Rightarrow \;\frac{1}{2} \le \cos (x) \le \frac{{\sqrt 2 }}{2}\)

 

[ku1005]

so can I just say that...or should I provide some proof because Im not quite sure how you came up with that???

like...its obvious to me in my head....but can I just say it??

 

[soroban]

Re: Using properties of Integrals

Hello, Rhys!

Show that: \(\displaystyle \L\:\frac{1}{96}\;\leq\; \frac{1}{\pi^2}\L\int^{\;\;\;\frac{\pi}{3}}_{\frac{\pi}{4}}x\cdot\cos x\,dx\;\leq\; \frac{\sqrt{2}}{36}\)

In the following, \(\displaystyle x\) is on the interval \(\displaystyle \left[\frac{\pi}{4},\,\frac{\pi}{3}\right]\)

Let \(\displaystyle \L\:I \;=\;\frac{1}{\pi^2}\int^{\;\;\;\frac{\pi}{3}}_{\frac{\pi}{4}}x\cdot\cos x\,dx\)


Conisder the right half of the inequality.

Since \(\displaystyle \,\cos x \:\leq \: 1\), then: \(\displaystyle x\cdot\cos x \:\leq \:x\)

Hence: \(\displaystyle \L\:I\;=\;\frac{1}{\pi^2}\int^{\;\;\;\frac{\pi}{3}}_{\frac{\pi}{4}}x\cdot\cos x\,dx \;\leq \;\frac{1}{\pi^2}\int^{\;\;\;\frac{\pi}{3}}_{\frac{\pi}{4}} x\,dx\)

. . We have: \(\displaystyle \L\:I \;\leq \;\int^{\;\;\;\frac{\pi}{3}}_{\frac{\pi}{3}}x\,dx\;=\;\frac{1}{\pi^2}\cdot\frac{1}{2}x^2\,]\begin{array}{cc}\frac{\pi}{3} \\ \\ \\ \frac{\pi}{4}\end{array} \;=\;\frac{7}{288} \;=\;0.024305...\)

. . Since \(\displaystyle \L\frac{\sqrt{2}}{36}\:=\:0.03928...\), we have: \(\displaystyle \L\:\frac{7}{288} \:<\:\frac{\sqrt{2}}{36}\)

Therefore: \(\displaystyle \L\:I \:< \:\frac{\sqrt{2}}{36}\)


For the left half of the inequality, we must show that: \(\displaystyle \L\:I \:\geq \:\frac{1}{96}\)

I'll let you work it out . . .

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\frac{x}{2} \le x\cos (x) \le \frac{{x\sqrt 2 }}{2} \\
\int\limits_{\pi /4}^{\pi /3} {\frac{x}{2}dx} \le \int\limits_{\pi /4}^{\pi /3} {x\cos (x)dx} \le \int\limits_{\pi /4}^{\pi /3} {\frac{{x\sqrt 2 }}{2}dx} \\
\end{array}\)