if ∫ f(x) dx = i ( UL= b, LL=a) ∫ f( x - (1/x)) dx ( UL=b,LL=a) =?
UL=upper limit
LL=lower limit
1let f(x)=x
then f(x - (1/x))=x-(1/x)
so now
∫lim a to b f(x) dx = i
changes to
∫lim a to b x dx = i
(b^2-a^2)/2=i
and
∫lim a to b f(x - (1/x))
=∫lim a to b x-(1/x)
=(x^2)/2-lnx|lim a to b
=(b^2-a^2)/2 + ln(a/b)
now (b^2-a^2)/2=i (from above)
so the answer is
i+ln(b/a)
3what if f(x) = lnx ........ dude arshad u have to generalise the ans. for any function
f(x) ...
1but mine is an assumption.........try doing it with any other function and please do tell me if u get the same answer............ and shriya what is the answer to this question........?
and my sole motive was to asume something which was easily integrable.......
