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Integrals giving me a hard time

  1. Apr 15, 2014 #1

    Rectifier

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    Gold Member

    Hey!
    It is the first time I post on this subform. Please forgive me if I do something wrong.

    1. The problem statement, all variables and given/known data
    [tex]F(x)=\int^x_0f(t)dt[/tex]
    for
    [tex]R \in t \rightarrow f(t) [/tex]

    2. Relevant equations
    Is it true that

    [tex]0 \leq f(x)\leq 3[/tex] for [tex]0<x<1[/tex]

    [tex]\int^x_0 tf(t)dt \leq x^2[/tex]

    for all [tex]x \in (0, 1)[/tex]? "


    3. The attempt at a solution
    [tex]\int^x_0 tf(t)dt \leq x^2 \Leftrightarrow t \int^x_0 f(t)dt \leq x^2 \Leftrightarrow t F(x) \leq x^2 [/tex]
    I dont know how to continue from here.
     
    Last edited: Apr 15, 2014
  2. jcsd
  3. Apr 15, 2014 #2
    If I'm understanding this correctly, you could start by writing [itex]x^{2}[/itex] as [itex]\int^{x}_{0}2tdt[/itex] and then comparing this to the original integral. Will the inequality be valid for any value of [itex]f(x)[/itex] on [itex]x\in (0,1)[/itex]?
     
  4. Apr 15, 2014 #3

    Rectifier

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    Gold Member

    Thank you for your reply!

    [tex] \int^x_0 tf(t) dt \leq \int^x_0 2t dt [/tex]

    Should I insert x=1(I cant insert x=0 since integrals must have bounds that are not equal)? What should I do with the f(t)?

    EDIT:
    Sorry, I just started this chapter and I am not so sure what to do.
     
    Last edited: Apr 15, 2014
  5. Apr 15, 2014 #4
    You could rewrite the inequality as

    [itex]\int^{x}_{0}[f(t)-2]tdt\leq0[/itex]

    Now, since [itex]x \in (0,1)[/itex], the [itex]t[/itex] will clearly not be negative on the interval, meaning [itex]f(t)-2[/itex] will either oscillate between positive and negative values, or will just be negative (since their product must be negative on at least one subset of (0,1) for the whole integral to be negative or zero). Now, since [itex]f(t)[/itex] can be ANY function (as long as [itex]f(t)\in(0,3)[/itex]), you can sometimes pick it to be the simplest one, in this case a constant. Can you then choose a constant f to simultaneously satisfy the above inequality and the initial assumption?
     
  6. Apr 15, 2014 #5

    Rectifier

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    Gold Member

    Thank you for your reply once again!

    I could pick f(t)=2 that will satisfy the inequality. But if I choose f(t)=3 it will mean that the inequality is wrong since the integral will be postitive. I guess that means that the initial inequality is wrong.

    Is that right?
     
  7. Apr 15, 2014 #6

    Mark44

    Staff: Mentor

    I don't understand what the above is supposed to mean. Can you explain in words what you're trying to convey here?
    The wording here is somewhat confusing, IMO. What part is given (or we are supposed to assume) and what part are we to show? IOW, are we supposed to verify that 0 ≤ f(x) ≤ 3 when 0 < x < 1?

    And then show that the inequality is true?
    You can't pull t outside the integral like you have done above.

    You are giv
     
  8. Apr 15, 2014 #7
    Yup, you found a counterexample, so the initial theorem doesn't hold, that's all there is to it.
     
  9. Apr 15, 2014 #8

    Rectifier

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    Gold Member

    Thank you for your post.
    I am having a hard time translating the problem to English from Swedish. We were supposed to verify if the inequality is true having (0 ≤ f(x) ≤ 3 when 0 < x < 1) as our starting conditions.
     
  10. Apr 15, 2014 #9

    Rectifier

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    Gold Member

    Thank you :)
     
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