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Integration Problem with a Variable for an Upper Limit

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

    If f(x) = integration from 0 to x of ( 1 / sqrt [ (t^3) + 2] ) dt , which of the following is FALSE?

    [tex]


    int({1}/{sqrt(t^3 + 2)},x=a..b)
    [/tex]

    a.) f(0) = 0
    b) f is continous at x for all x >= 0
    c) f(1) > 0
    d) f'(1) = 1 / sqrt [3]
    e) f(-1) > 0.

    I know how to do a,b, and d. But how can I test for c and d if I don't know how to integrate such a weird function? I tried thinking of an arctan integration.... but it looks "weird".

    Thanks
     
    Last edited: Feb 24, 2009
  2. jcsd
  3. Feb 24, 2009 #2

    Dick

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    You don't necessarily have to integrate a function to know something about it's values. For t in [-1,1], what is the sign of 1/sqrt(t^3+2)?
     
  4. Feb 24, 2009 #3
    I can see that the graphof 1 / sqrt(t^3 + 2) from [-1,1] has a positive area. However, is this enough to prove that the integration is going to be positive?

    I graphed the function... and when we take the integration from 0 to -1... will yield a negative error. Therefore, e) is wrong, correct?
     
  5. Feb 24, 2009 #4

    Dick

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    Right. And, yes, if you have a positive area, the integral is positive. The integral measures area.
     
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