definite integration

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definite integration

Mathematics
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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

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well, wolfram used a gamma function; but i got no idea what that is :) the answer they gave is: http://www4c.wolframalpha.com/Calculate/MSP/MSP41119fe0hefdegbcb8d000036e7b90hdib4d907?MSPStoreType=image/gif&s=5&w=188&h=49
its greater than pi/4 tho :)
\(\sqrt{1+x^4}>\sqrt{1+x^2}>\sqrt{1-x^2}\) Then the integral is greater than \(\int_0^1\sqrt{1-x^2}\,dx=\text{one quarter of a circle}=\pi/4\) So \(I>\pi/4\).
Oh sorry, I miss read the problem :(
But the idea is the same. Notice that \(0\leq (x-x^3)^2=x^2+x^6-2x^4\) Hence \(x^2+x^6\geq 2x^4\geq x^4.\) Now \(\sqrt{1+x^4}\sqrt{1-x^2}=\sqrt{1+x^4-x^2-x^6}< 1\) for \(x>0\) Hence \(\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\) Hence we can use the same argument as above to show that \(I>\pi/4\).
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
:) To read a solution you need to read it backward sometimes. Obviously we want to show that \(\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\). The inequality is equivalent to \(\sqrt{1+x^4-x^2-x^6}<1\) and I can show that if know that \(x^4-x^2-x^6<0\) and luckily I can prove the last inequality from \((x-x^3)^2\geq 0\).
ok i understand most of it. but can you tell me once what led you to show this: \[\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\]
watchmath need ur help again sorry about this
Because I know (from my past experience) that the integral \(\int_0^1\sqrt{1-x^2}\,dx\) represent a quarter of the area of a circle with radius 1. Sketch the graph of \(y=\sqrt{1-x^2}\) and see what is the area under the graph from \(x=0\) to \(x=1\).
okay its pi/4... but the doubt persists..:-(
what doubt?
if integrating \[\sqrt{1-x^2}\] gives pi/4 then how does integrating \[1/\sqrt{1+x ^{4}}\] give a value greater than pi/4
Well remember that if the graph of \(f\) is above the graph of \(g\) then the area under the curve of \(f\) is greater than the area under the graph of \(g\). Now for \(f=1/\sqrt{1+x^4}\) and \(g=\sqrt{1-x^2}\) we have \(f>g\). So the integral (the area under the \(f\)) is bigger then the area under the graph of \(g\). But the area under the graph of \(g\) is \(\pi/4\).So \(\int_0^1 f>\pi/4\).
ok hold on.. if we know " that if the graph of f is above the graph of g then the area under the curve of f is greater than the area under the graph of g." then we dont need to calculate anything right?? that itself answers the question.
how will you answer the question if you per se dont know that the graph of f is greater than that of g
Well first we need to make a reasonable hypothesis. For example after observing the the graph of the above \(f\) and \(g\) (by sketching the two in calculator)we believe that \(fg\), then we try to prove that rigorously. When you see a theorem or a solution to a problem you won't see many untold story behind that proof/solution. It is possible before arrive to that solution, the author revise the hypothesis, try a lot of things that doesn't work before he can write a neat and beautiful solution. In this problem for example we have three choices: the integral is less, equal or greater than pi/4. the beginning that we believe that the answer is pi/4. But after several attempts we might see that maybe this is not true. Then we try to show that it is less that pi/4 but maybe realize after many attempts that it is unlikely to be true. Then we try to prove it greatern that pi/4. And the above is one possible scenario how to come up with the prove for that.
hmm....i see what your point is.

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