anonymous
  • anonymous
definite integration
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits_{0}^{1}dx/\sqrt{1+x ^{4}}=I\] then a) I>pi/4 b)I
amistre64
  • amistre64
\[\int\limits_{0}^{1} \frac{dx}{\sqrt{1+x^4}}\]
anonymous
  • anonymous
yes

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amistre64
  • amistre64
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
amistre64
  • amistre64
its greater than pi/4 tho :)
watchmath
  • watchmath
\(\sqrt{1+x^4}>\sqrt{1+x^2}>\sqrt{1-x^2}\) Then the integral is greater than \(\int_0^1\sqrt{1-x^2}\,dx=\text{one quarter of a circle}=\pi/4\) So \(I>\pi/4\).
watchmath
  • watchmath
Oh sorry, I miss read the problem :(
watchmath
  • watchmath
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 \(I>\pi/4\).
anonymous
  • anonymous
how did you get this watchmath\[0\leq (x-x^3)^2=x^2+x^6-2x^4\]
watchmath
  • watchmath
Fir every square is non-negative and the rest you just expand \((x-x^3)(x-x^3)\)
anonymous
  • anonymous
no thats alright. why (x-x^3)^2??
watchmath
  • watchmath
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.
anonymous
  • anonymous
m sorry bt m rly not gettin it.. why do you wanna show x^2+x^6-x^4> 0
anonymous
  • anonymous
watchmath i need ur help again look at my comment i made on my question please
watchmath
  • watchmath
:) 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 \((x-x^3)^2\geq 0\).
anonymous
  • anonymous
ok i understand most of it. but can you tell me once what led you to show this: \[\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\]
anonymous
  • anonymous
watchmath need ur help again sorry about this
watchmath
  • watchmath
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 \(x=1\).
anonymous
  • anonymous
okay its pi/4... but the doubt persists..:-(
watchmath
  • watchmath
what doubt?
anonymous
  • anonymous
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
watchmath
  • watchmath
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 \(\int_0^1 f>\pi/4\).
anonymous
  • anonymous
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.
anonymous
  • anonymous
how will you answer the question if you per se dont know that the graph of f is greater than that of g
watchmath
  • watchmath
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.
anonymous
  • anonymous
hmm....i see what your point is.

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