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Integration using given values

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Let f(x) be a continuous function defined on the interval [2, ∞) such that

    f(2) = 12

    |f(x)| < x^8 + 8

    and

    ⌠∞
    f(x) * e^(−x/4) dx = −8
    ⌡2

    Determine the value of

    ⌠∞
    f′(x) * e^(−x/4) dx
    ⌡2


    2. Relevant equations



    3. The attempt at a solution

    I started by writing out the integrations. So I have

    -4F(∞)e^(-∞/4) + 4(F(2)*e^(-1/2)) = -8
    and
    -4f(∞)e^(-∞/4) + 4f(2)*e^(-1/2) = ?

    I see that I can use f(2) = 12 in the second equation, but I don't see how to use the first equation at all. Nor do I see how |f(x)| < x^8 + 8 can help me. Thoughts?
     
  2. jcsd
  3. Oct 18, 2007 #2
    I'm not terrible sure why you need

    [tex] |f(x)| < x^8 + 8 [/tex]. It seems a bit too specific since we can get an answer without needing that level of specificity (unless I'm wrong in my solution)


    All it depends on is the fact that [tex]O(f(x)) = x^n [/tex]

    Use integration by parts on the equation you're trying to find a value for. Then you will use all the information you've been given.

    Note: you will need [tex] |f(x)| < x^8 + 8 [/tex]. to justify why

    [tex] e^{\frac{-x}{4}} f(x) \rightarrow 0 [/tex] as [tex] x\rightarrow \infty [/tex]
     
  4. Oct 18, 2007 #3
    Thank you! Everything works out perfectly after that.

    So after integration by parts, I have:

    I = f(x)*e^(-x/4) + (1/4)int f(x)*e^(-x/4)dx

    Plug in the values and I have:

    f(∞)e^(-∞/4) - 12e^(-1/2) + (1/4)(-8)

    0 -12e^(-1/2) -2
     
    Last edited: Oct 18, 2007
  5. Oct 18, 2007 #4
    Right, but don't forget to evaluate at the limits of integration, and note why we do actually need

    [tex] |f(x)| < x^8 + 8 [/tex]

    though not in that level of specificity. The problem could've stated that as any polynomial for it to work under the method I've shown here. If you don't know enough about big-oh notation to note that
    [tex] e^{\frac{-x}{4}} f(x) \rightarrow 0 [/tex] as [tex] x\rightarrow \infty [/tex]

    then try and convince yourself by taking the limit of the product as [tex] x\rightarrow \infty [/tex]. Note that you will have to use L'Hopitals rule, both factors are infinitely differentiable, and that [tex]\forall p_n(x) \in P_n[x] \: \frac{d^{n+1}}{dx^{n+1}}p_n(x) = 0[/tex]
     
    Last edited: Oct 18, 2007
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