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Math Help - Integral question

  1. #1
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    Integral question

    Hi,
    I'm having some trouble with this question. I've made a start on it but I'm not sure how to finish it.

    Let f be differentiable in [a,b] and let f' be continuous in [a,b]. f(a)=0. Prove the inequality in the attached image.

    What I did was deal with the left side:
    LHS = (b-a)(|f(b) - f(a)|) = (b-a)|f(b)|

    But I don't understand why this is necessarily bigger than the RHS.

    Thanks
    Attached Thumbnails Attached Thumbnails Integral question-2c.jpg  
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  2. #2
    Super Member girdav's Avatar
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    Re: Integral question

    We have f(t)-f(a) =\int_a^t f'(s)ds hence |f(t)|\leq \int_a^t|f'(s)|ds\leq \int_a^b |f'(s)|ds. Now take the integral on both sides.
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  3. #3
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    Re: Integral question

    I don't understand what you've done or how it answers the question.
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  4. #4
    Super Member girdav's Avatar
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    Re: Integral question

    I only used the fundamental theorem of calculus. Now if you follow my hint the result is shown.
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  5. #5
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    Re: Integral question

    Quote Originally Posted by durrrrrrrr View Post
    I don't understand what you've done or how it answers the question.
    Do you agree that f(t) = \int_a^t {f'(s)ds}~?

    If so, then using basic theorems we get
    \left| {f(t)} \right| = \left| {\int_a^t {f'(s)ds} } \right| \leqslant \int_a^t {\left| {f'(s)} \right|ds}  \leqslant \int_a^b {\left| {f'(s)} \right|ds}

    Here it last trick: B=\int_a^b {\left| {f'(s)} \right|ds} is a number.

    Thus \int_a^b {\left| {f(x)} \right|dx}  \leqslant \int_a^b {Bdx = \left( {b - a} \right)B}
    Last edited by Plato; July 11th 2011 at 01:05 PM.
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