1i m also geting wht u r getting eureka...may b zero becoause in definite integration constant is 0......
f(1/x)+x2f(x)=0 for all x>0
and I=\int_{1/x}^{x}{f(z)dz}
1/2<=x<2
i did like this
f(1/x)+x2f(x)=0 ----------(1)
replace x by 1/x
=>f(x)+1/x2 f(1/x)=0
Now dI/dx=f(x).1 -(-1/x2 f(1/x))=> dI/dx=f(x)+1/x2 f(1/x)
dI/dx=0
=>dI=0
=>I=constant
but answer is I=0
where is mistake ??
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12 Answers
1f(1/x)+x2f(x)=0 ----------(1)
replace x by 1/x
=>f(x)+1/x2 f(x)=0...here is the mistake...it will be =>f(x)+1/x2 f(1/x)=0..u fogot that u replaced x by 1/x here
24oh hthats typo..i forgot to write it here...i have done tha in my notebook
11even if I = c
put x = 1 in I = .....
u will get c= 0 ( integral 1 to 1 f(z) is zero)
1es m right..u cannot have a constant in a definite integral if the initial n final valus are x and 1/x or wateva function of x......so the answeris indeed zero
3also eure u dont know the value of that constant it can also be zero na.
i havent done in a paper but i think u have to substitute t=1/x in the integral equation.
use the functional eqn given.
1I = ∫( from 1/x to x) f(z) dz = ∫( from 1/x to x) ( -1/x2 f(1/x) ) dx ( from d condition )
now, let 1/x = m so, then : [ - 1/x2 dx = dm ]
change the limits. now,it becomes : ∫( from x to 1/x) f(m) dm.
= ∫(from x to 1/x) f(z) dz . now, change the limits : - ∫(from 1/x to x) f(z) dz = ( - I )
SO, I = -I .THUS, I = 0.
1eure,watch out the properties used here.
first step ∫( from 1/x to x) f(z) dz = ∫ (from 1/x to x) f(x) dx
2nd step wrote f(1/x) + x2f(x)=0 for all x>0 , as : f(x) = f(1/x) * (-1/x2). then substituted 1/x=m.
3rd step changed the limits . limits got interchanged. thus i changed the limits again and put a minus sign.
last step used the property of first step and again replaced : - ∫( 1/x to x) f(m) dm = - ∫( from 1/x to x) f(z) dz
now, I = -I .THUS, I = 0.
hope got it clear now.