i really need to know....by moham_87 Tags: None 

#1
Jan1004, 03:30 PM

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 



#2
Jan1004, 03:43 PM

P: 661

For first part
[tex]\int \frac{x dx}{x^4+2x^2+1} = \int \frac{x dx}{(x^2+1)^2}[/tex] now substitute x^2+1=t u will get 2xdx=dt and so on u get [tex]\int \frac{dt}{2t^2}=\frac{1}{2t}[/tex] 



#3
Jan1004, 03:46 PM

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 :) 



#4
Jan1004, 04:02 PM

P: 661

i really need to know.... 



#5
Jan1004, 04:06 PM

P: 201

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 



#6
Jan1004, 04:27 PM

P: 661

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] 



#7
Jan1004, 06:00 PM

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.




#8
Jan1004, 06:38 PM

P: 211





#9
Jan1004, 06:51 PM

P: 201

and mmwave as I told you go to sosmath and there you find lots of students who can share math ideas with you :) 



#10
Jan1004, 10:44 PM

P: 678

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. 



#11
Jan1004, 11:02 PM

P: 46





#12
Jan1004, 11:35 PM

P: 678




Register to reply 