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anonymous

  • 5 years ago

definite integration

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{1}dx/\sqrt{1+x ^{4}}=I\] then a) I>pi/4 b)I<pi/4 c)i=pi/4 d)none of these

  2. amistre64
    • 5 years ago
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    \[\int\limits_{0}^{1} \frac{dx}{\sqrt{1+x^4}}\]

  3. anonymous
    • 5 years ago
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    yes

  4. amistre64
    • 5 years ago
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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

  5. amistre64
    • 5 years ago
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    its greater than pi/4 tho :)

  6. watchmath
    • 5 years ago
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    \(\sqrt{1+x^4}>\sqrt{1+x^2}>\sqrt{1-x^2}\) Then the integral is greater than \(\int_0^1\sqrt{1-x^2}\,dx=\text{one quarter of a circle}=\pi/4\) So \(I>\pi/4\).

  7. watchmath
    • 5 years ago
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    Oh sorry, I miss read the problem :(

  8. watchmath
    • 5 years ago
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    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 \(I>\pi/4\).

  9. anonymous
    • 5 years ago
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    how did you get this watchmath\[0\leq (x-x^3)^2=x^2+x^6-2x^4\]

  10. watchmath
    • 5 years ago
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    Fir every square is non-negative and the rest you just expand \((x-x^3)(x-x^3)\)

  11. anonymous
    • 5 years ago
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    no thats alright. why (x-x^3)^2??

  12. watchmath
    • 5 years ago
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    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.

  13. anonymous
    • 5 years ago
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    m sorry bt m rly not gettin it.. why do you wanna show x^2+x^6-x^4> 0

  14. anonymous
    • 5 years ago
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    watchmath i need ur help again look at my comment i made on my question please

  15. watchmath
    • 5 years ago
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    :) 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 \((x-x^3)^2\geq 0\).

  16. anonymous
    • 5 years ago
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    ok i understand most of it. but can you tell me once what led you to show this: \[\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\]

  17. anonymous
    • 5 years ago
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    watchmath need ur help again sorry about this

  18. watchmath
    • 5 years ago
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    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 \(x=1\).

  19. anonymous
    • 5 years ago
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    okay its pi/4... but the doubt persists..:-(

  20. watchmath
    • 5 years ago
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    what doubt?

  21. anonymous
    • 5 years ago
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    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

  22. watchmath
    • 5 years ago
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    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 \(\int_0^1 f>\pi/4\).

  23. anonymous
    • 5 years ago
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    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.

  24. anonymous
    • 5 years ago
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    how will you answer the question if you per se dont know that the graph of f is greater than that of g

  25. watchmath
    • 5 years ago
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    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 \(f<g\) 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.

  26. anonymous
    • 5 years ago
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    hmm....i see what your point is.

  27. anonymous
    • 5 years ago
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    any ideas on this question http://openstudy.com/groups/mathematics/updates/4dd42d14d95c8b0b0f5156c4#/groups/mathematics/updates/4dd58012d95c8b0bcd095bc4

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