# proof, integrals

• May 24th 2013, 10:51 AM
ryu1
proof, integrals
Hi,

Need to prove
A. integral from pi to 2pi of (|cosx|)/(x^2) dx <= 1/2pi

B. 1/4 <= integral from 0 to 1 of (x+x^3)/(1+x+x^5) dx <= 3/4

Need to use the theorem that says:
1. If f is integable(has an integral) in [a,b] and for every x there is f(x) => 0
then
integral from a to b of f(x) => 0

2. if f and g are integrable in [a,b] and for every x there exists
f(x) => g(x)
then
the integral from a to b of f(x) dx => the integral from a to b of g(x) dx

• May 24th 2013, 11:06 AM
Plato
Re: proof, integrals
Quote:

Originally Posted by ryu1
prove
A. integral from pi to 2pi of (|cosx|)/(x^2) dx <= 1/2pi
B. 1/4 <= integral from 0 to 1 of (x+x^3)/(1+x+x^5) dx <= 3/4

Both of these are done the same way.

a) $\displaystyle \int_\pi ^{2\pi } {\frac{{\left| {\cos (x)} \right|dx}}{{{x^2}}}} \leqslant \int_\pi ^{2\pi } {\frac{{dx}}{{{x^2}}}}$
• May 24th 2013, 01:11 PM
ryu1
Re: proof, integrals
amazing, I have been breaking my head for a while with this...
although your explanation wasnt complete as usual it helped me see the point in my teacher lesson :)

Thank you once again.

My "proof" (according to my teacher "template" of another example) :

in the interval pi<= x <= 2pi

|cos| <= 1

multiplying by 1/x^2 we get
|cos|/x^2 <= 1/x^2

according to the 5.6... theorem (it's called statement 5.6 something in the textbook, thats how he wants us to write the proof)
from that we derive Plato's answer (the integral of the first function (with the |cos|) is <= the integral of 1/x^2 in the interval [pi,2pi] because |cos|/x^2 <= 1/x^2)

solving the integral of 1/x^2 in [pi,2pi] we get 1/2pi

therefore the integral of |cos|/x^2 dx in [pi,2pi] is <= 1/2pi

so cool!

Now for the second one, you say it can be solved the same way... the teacher actually wrote some hint in the homework instructions for this one.
I'll see if I can solve it.

Thanks!

P.S
How do you write the integral sign and everything so neatly here?
• May 24th 2013, 02:00 PM
Plato
Re: proof, integrals
Quote:

Originally Posted by ryu1
B. 1/4 <= integral from 0 to 1 of (x+x^3)/(1+x+x^5) dx <= 3/4

Can you show that:
for $\displaystyle x\in (0,1)$ then $\displaystyle \frac{x+x^3}{3}\le \frac{x+x^3}{1+x+x^5}\le x+x^3~?$
• May 24th 2013, 02:43 PM
ryu1
Re: proof, integrals
yes, I actually just finished solving the second problem.

in interval [0,1] ,

for right side x+x^5+1 => 1+0+0 taking the reciprocal we get 1/1+x+x^5 <= 1
multiply by x+x^3 we get x+x^3/1+x+x^5 <= x+x^3

for left side, 3 => 1+x+x^5, taking the reciprocal we get 1/3 <= 1/x+x^5, multiply by x+x^3 , x+x^3/3 <= x+x^3/1+x+x^5

It's the way I used (and was expected) to use to explain the original problem inequality.

how do you write equations with all the proper signs in the forum?

Thanks.
• May 24th 2013, 04:47 PM
Plato
Re: proof, integrals
Quote:

Originally Posted by ryu1
how do you write equations with all the proper signs in the forum?

You must learn to use LaTeX.
It really is so easy. And it makes most of us more willing to find out how to help.
This subforum will help you with the code. Once you begin, you quickly learn the code.

[TEX]\int_\pi ^{2\pi } {\frac{{\left| {\cos (x)} \right|dx}}{{{x^2}}}} \leqslant \int_\pi ^{2\pi } {\frac{{dx}}{{{x^2}}}} [/TEX]

[TEX]x\in (0,1)[/TEX] then [TEX]\frac{x+x^3}{3}\le \frac{x+x^3}{1+x+x^5}\le x+x^3~?[/TEX]

Gives $\displaystyle \int_\pi ^{2\pi } {\frac{{\left| {\cos (x)} \right|dx}}{{{x^2}}}} \leqslant \int_\pi ^{2\pi } {\frac{{dx}}{{{x^2}}}}$

$\displaystyle x\in (0,1)$ then $\displaystyle \frac{x+x^3}{3}\le \frac{x+x^3}{1+x+x^5}\le x+x^3~?$[
• May 24th 2013, 06:00 PM
ryu1
Re: proof, integrals
ok I'll test it in my next thread.
Thanks.