1. ## 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

2. ## Re: proof, integrals

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

3. ## 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?

4. ## Re: proof, integrals

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 $x\in (0,1)$ then $\frac{x+x^3}{3}\le \frac{x+x^3}{1+x+x^5}\le x+x^3~?$

5. ## 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.

6. ## Re: proof, integrals

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

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

7. ## Re: proof, integrals

ok I'll test it in my next thread.
Thanks.