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Proof for this inequality

  1. Nov 29, 2015 #1
    • Member warned about posting with no effort shown
    1. The problem statement, all variables and given/known data
    iqitd3.jpg

    2. Relevant equations
    With the regards to posting such a incomplete equation, I will soon put in the updated one
    Thank you

    3. The attempt at a solution
    visual graph... didn't help
     
    Last edited: Nov 29, 2015
  2. jcsd
  3. Nov 29, 2015 #2

    pasmith

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    What is the range of integration?
     
  4. Nov 29, 2015 #3
    Our teacher did not write any range :/
     
  5. Nov 29, 2015 #4

    Ray Vickson

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    PF Rules require you to make some efforts and to show your work before seeking help (or, more precisely, they do not permit us to help until you have shown your work).

    Anyway, the question is incomplete: what are the integration limits?

    Note added in edit: I see that the latter question has already been posed to you, but appeared on my screen only after I pressed the 'enter' key.
     
  6. Nov 29, 2015 #5
    Certainly I didn't tackle the equation without effort but I didn't want to send my rather nonsensical and messy attempts .
    I replied the last question you asked above
     
  7. Nov 29, 2015 #6

    Ray Vickson

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    If you do not show your "nonsensical and messy" attempts, how are we supposed to know where you are having trouble? We need to see what you have done, wrong or not! (BTW: I edited my previous post, which dealt with your response about integration limits).
     
  8. Nov 29, 2015 #7
    Okay haha!
    Well I looked for the first function which turned out a complex function. and I graphed the function using the Algeo app but still didn't help. I tried setting the limits pi fourth to half pi but still is risky
    Our teacher suggested a guide tip for using the Mean value theorem... But it doesn't have a range !!!!
     
  9. Nov 29, 2015 #8

    micromass

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    How can you even begin to solve this if you don't know the integration limits???? This is crucial information
     
  10. Nov 29, 2015 #9

    Ray Vickson

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    Those limits cannot possibly work, as they would involve values of ##x## for which ##\sin^{-1} (x)## is either undefined, or not real. (I am assuming you mean the inverse sine function, rather than the reciprocal ##1/\sin x##.)
     
  11. Nov 29, 2015 #10
    rather should I storm upon the teacher for this crucial mistake?
     
  12. Nov 29, 2015 #11

    micromass

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  13. Nov 29, 2015 #12
    definitely I mean the inverse sine function. Well it has come to a conclusion that limits are needed and essential
     
  14. Nov 29, 2015 #13
    Thanks for all the help
     
  15. Nov 30, 2015 #14

    Ray Vickson

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    If ##f(x) = \arcsin(x)/(1+x^8)##, the possible case ##\int_{-1}^1 f(x) \, dx## is easy to compute explicitly (with almost no work), and the left-hand inequality is FALSE but the right-hand one is (trivially) true. Thus, if the integration limits are -1 and +1, only one of the two inequalities is true.

    The other obvious possibility, ##\int_0^1 f(x) \, dx##, takes more work; both bounds are true, as shown by numerical evaluation of the integral. However, proving it "analytically" is a challenge.
     
    Last edited: Nov 30, 2015
  16. Nov 30, 2015 #15
    I did try this today in fact. Tomorrow I will see my Math teacher and definitely will ask why she made this mistake and see what she had in mind!
    Thank you for your help dear friend
     
  17. Nov 30, 2015 #16

    Ray Vickson

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    Please note that I made an error in post #14 (due to a typo). I have edited that post, and you should look at it again.
     
  18. Dec 1, 2015 #17
    I come updated.
    The limits are 0 to positive 1
     
  19. Dec 1, 2015 #18
     
  20. Dec 1, 2015 #19

    fresh_42

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  21. Dec 1, 2015 #20

    SammyS

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    For the upper bound, compare to ##\displaystyle \ \int_0^1 {\sin^{-1\,}\!(x)}\ dx \ ##.
     
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