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\[\int\limits_{0}^{1}dx/\sqrt{1+x ^{4}}=I\] then
a) I>pi/4
b)I

\[\int\limits_{0}^{1} \frac{dx}{\sqrt{1+x^4}}\]

yes

its greater than pi/4 tho :)

Oh sorry, I miss read the problem :(

how did you get this watchmath\[0\leq (x-x^3)^2=x^2+x^6-2x^4\]

Fir every square is non-negative and the rest you just expand \((x-x^3)(x-x^3)\)

no thats alright. why (x-x^3)^2??

Because I want to show that \(x^2+x^6-x^4> 0\) and we can use \((x-x^3)^2\geq 0\) to prove that.

m sorry bt m rly not gettin it..
why do you wanna show x^2+x^6-x^4> 0

watchmath i need ur help again look at my comment i made on my question please

watchmath need ur help again sorry about this

okay its pi/4...
but the doubt persists..:-(

what doubt?

hmm....i see what your point is.