# i really need to know....

by moham_87
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 P: 13 first-> integration of: x/(x^4 + 2x^2 + 1) i tried solving that using (ln), u=g(x), and alot of ways, but i couldn't get an answer second-> ----------------------------------------------------------------- Is the following argument valid? Explain f(x) = 1 if (x) is rational and (x)>0 -1 if (x) is rational and (x)<0 0 if (x) is irrational is defined for all numbers in [-1,1] and has the property that f(-x)=-f(x) for all (x) in [-1,1]. Thus, definite integration{0 to 1 of f(x).dx}=0 ----------------------------------------------------------------- i couldn't understand the question, and what is meant by "irrational", i think all values (even integers) can be rational thank u alot, i wish to reply as soon as possible, my exam is TOMORROW.... and any efforts will be appreciated
 P: 661 For first part $$\int \frac{x dx}{x^4+2x^2+1} = \int \frac{x dx}{(x^2+1)^2}$$ now substitute x^2+1=t u will get 2xdx=dt and so on u get $$\int \frac{dt}{2t^2}=-\frac{1}{2t}$$
 P: 201 Man for Exams and these types of math go and consult: http://www.sosmath.com/CBB/index.php I think here you'll find much help :)
P: 661

## i really need to know....

 Man for Exams and these types of math go and consult:
I believe it should be in Homework Section And nothing else[:))]
P: 201
 Originally posted by moham_87 second-> ----------------------------------------------------------------- Is the following argument valid? Explain f(x) = 1 if (x) is rational and (x)>0 -1 if (x) is rational and (x)<0 0 if (x) is irrational is defined for all numbers in [-1,1] and has the property that f(-x)=-f(x) for all (x) in [-1,1]. Thus, definite integration{0 to 1 of f(x).dx}=0 ----------------------------------------------------------------- i couldn't understand the question, and what is meant by "irrational", i think all values (even integers) can be rational thank u alot, i wish to reply as soon as possible, my exam is TOMORROW.... and any efforts will be appreciated [/B]
A number that can be expressed as a fraction where p and q are integers and , is called a rational number with numerator p and denominator q. Numbers that are not rational are called irrational numbers.

For the validness...I am not sure...first you've to define the function!

But as the function seems odd, of course the integration will be 0
P: 661
 A number that can be expressed as a fraction where p and q are integers and , is called a rational number with numerator p and denominator q. Numbers that are not rational are called irrational numbers.
Nonperiodic decimal fractions are called irrational numbers.

 But as the function seems odd, of course the integration will be 0
It would be zero only when integration is of the form $$\int^a_{-a}$$

For the second part

Any number on x axis has two possibility either rational OR irrational for rational f(x)=1 whereas fo irational it is = 0,

A rational point is surrounded by two irrational points and vice versa So clearly the function is discontinuous

And function consists of a collection of points as defined by function

integral wont be zero in this range i.e from [0,1]
 P: 678 The second function is definitely not integrable. On the interval [0,1] any upper sum will be 1 and any lower sum will be 0. Thus there is no value for the integral.
P: 211
 Originally posted by himanshu121 [B] A rational point is surrounded by two irrational points and vice versa So clearly the function is discontinuous
This is a fascinating tidbit I don't recall learning. Is it taught in calculus? I'd like to learn more about this idea, maybe see a proof if it isn't too complicated. Can you suggest a reference?
P: 201
 Originally posted by himanshu121 Nonperiodic decimal fractions are called irrational numbers. It would be zero only when integration is of the form $$\int^a_{-a}$$ For the second part Any number on x axis has two possibility either rational OR irrational for rational f(x)=1 whereas fo irational it is = 0, A rational point is surrounded by two irrational points and vice versa So clearly the function is discontinuous And function consists of a collection of points as defined by function integral wont be zero in this range i.e from [0,1]
Thanks! Himanda! I read once and didn't watched the limits :(

and mmwave as I told you go to sosmath and there you find lots of students who can share math ideas with you :)
P: 678
 Originally posted by mmwave This is a fascinating tidbit I don't recall learning. Is it taught in calculus? I'd like to learn more about this idea, maybe see a proof if it isn't too complicated. Can you suggest a reference?
Suppose that you have a rational point not surrounded by two irrational points. So you have two rational numbers "a" and "b" with no irrational number between them. Thus the interval [a,b] contains only rational numbers, even over the reals.

But we clearly have a bijection between [a,b] and [0,1] (just apply a translation and scale the interval). Thus both [a,b] and [0,1] have the same cardinality.

Yet [a,b] only contains rational numbers and so it must be countable. But we can show that [0,1] is uncountable. Thus there cannot exist a bijection between [a,b] and [0,1]. This is a contradiction. Thus every rational is surrounded by two irrationals.
P: 46
 The second function is definitely not integrable. On the interval [0,1] any upper sum will be 1 and any lower sum will be 0. Thus there is no value for the integral.
The function is not Riemann integrable. But there are other sorts of integration, namely Lebesgue integration, that can handle such functions.
P: 678
 Originally posted by BigRedDot The function is not Riemann integrable. But there are other sorts of integration, namely Lebesgue integration, that can handle such functions.
I didn't want to confuse the original poster too much. I figured bringing up Lebesgue integration (or other types of integration) would do that.