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anonymous
 5 years ago
definite integration
anonymous
 5 years ago
definite integration

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}dx/\sqrt{1+x ^{4}}=I\] then a) I>pi/4 b)I<pi/4 c)i=pi/4 d)none of these

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1} \frac{dx}{\sqrt{1+x^4}}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its greater than pi/4 tho :)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1\(\sqrt{1+x^4}>\sqrt{1+x^2}>\sqrt{1x^2}\) Then the integral is greater than \(\int_0^1\sqrt{1x^2}\,dx=\text{one quarter of a circle}=\pi/4\) So \(I>\pi/4\).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Oh sorry, I miss read the problem :(

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1But the idea is the same. Notice that \(0\leq (xx^3)^2=x^2+x^62x^4\) Hence \(x^2+x^6\geq 2x^4\geq x^4.\) Now \(\sqrt{1+x^4}\sqrt{1x^2}=\sqrt{1+x^4x^2x^6}< 1\) for \(x>0\) Hence \(\frac{1}{\sqrt{1+x^4}}>\sqrt{1x^2}\) Hence we can use the same argument as above to show that \(I>\pi/4\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get this watchmath\[0\leq (xx^3)^2=x^2+x^62x^4\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Fir every square is nonnegative and the rest you just expand \((xx^3)(xx^3)\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no thats alright. why (xx^3)^2??

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Because I want to show that \(x^2+x^6x^4> 0\) and we can use \((xx^3)^2\geq 0\) to prove that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0m sorry bt m rly not gettin it.. why do you wanna show x^2+x^6x^4> 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath i need ur help again look at my comment i made on my question please

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1:) To read a solution you need to read it backward sometimes. Obviously we want to show that \(\frac{1}{\sqrt{1+x^4}}>\sqrt{1x^2}\). The inequality is equivalent to \(\sqrt{1+x^4x^2x^6}<1\) and I can show that if know that \(x^4x^2x^6<0\) and luckily I can prove the last inequality from \((xx^3)^2\geq 0\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i understand most of it. but can you tell me once what led you to show this: \[\frac{1}{\sqrt{1+x^4}}>\sqrt{1x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath need ur help again sorry about this

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Because I know (from my past experience) that the integral \(\int_0^1\sqrt{1x^2}\,dx\) represent a quarter of the area of a circle with radius 1. Sketch the graph of \(y=\sqrt{1x^2}\) and see what is the area under the graph from \(x=0\) to \(x=1\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay its pi/4... but the doubt persists..:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if integrating \[\sqrt{1x^2}\] gives pi/4 then how does integrating \[1/\sqrt{1+x ^{4}}\] give a value greater than pi/4

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Well 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 \(g=\sqrt{1x^2}\) 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 \(\int_0^1 f>\pi/4\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how will you answer the question if you per se dont know that the graph of f is greater than that of g

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Well 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 \(f<g\) or \(f>g\), 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm....i see what your point is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0any ideas on this question http://openstudy.com/groups/mathematics/updates/4dd42d14d95c8b0b0f5156c4#/groups/mathematics/updates/4dd58012d95c8b0bcd095bc4
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