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Thread: show its integral less than half M

  1. #1
    Senior Member
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    show its integral less than half M

    suppose
    f: R--->R has a continuous derivative, f(0)=0, and |f'(x)|<=M for x in [0,1]

    show that, |integral(f,0,1)|<= half M

    show that, if given f(1)=0, then |integral(f,0,1)|<= quarter M

    what could you say if |f'(x)|<= Mx?
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by szpengchao View Post
    suppose
    f: R--->R has a continuous derivative, f(0)=0, and |f'(x)|<=M for x in [0,1]

    show that, |integral(f,0,1)|<= half M

    show that, if given f(1)=0, then |integral(f,0,1)|<= quarter M

    what could you say if |f'(x)|<= Mx?
    $\displaystyle
    \left| {f'\left( x \right)} \right| \leqslant M \Leftrightarrow - M \leqslant f'\left( x \right) \leqslant M
    $

    Integrating: $\displaystyle
    - M \leqslant f'\left( x \right) \leqslant M \Rightarrow \int_0^t { (- M)dx} \leqslant \int_0^t {f'\left( x \right)dx} \leqslant \int_0^t {Mdx}
    $ whenever $\displaystyle
    0 \leqslant t \leqslant 1
    $

    By the FTC: $\displaystyle
    \int_0^t {f'\left( x \right)dx} = f\left( t \right) - f\left( 0 \right)
    $

    Thus: $\displaystyle
    \left( { - M} \right) \cdot t \leqslant f\left( t \right) - f\left( 0 \right) \leqslant M \cdot t

    $

    Integrating again: $\displaystyle
    \left( { - M} \right) \cdot \int_0^1 {t \cdot dt} \leqslant \int_0^1 {f\left( t \right) \cdot dt} - \int_0^1 {f\left( 0 \right) \cdot dt} \leqslant M \cdot \int_0^1 {t \cdot dt}
    $

    Therefore: $\displaystyle
    \left( {\tfrac{{ - M}}
    {2}} \right) \leqslant \int_0^1 {f\left( t \right) \cdot dt} - f\left( 0 \right) \leqslant \left( {\tfrac{M}
    {2}} \right)
    $

    Since $\displaystyle f(0)=0$ we get the desired result

    All the other parts are similar, try to do them and then ask if you still have doubts.

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  3. #3
    Senior Member
    Joined
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    how can u get the second ques?

    Quote Originally Posted by PaulRS View Post
    $\displaystyle
    \left| {f'\left( x \right)} \right| \leqslant M \Leftrightarrow - M \leqslant f'\left( x \right) \leqslant M
    $

    Integrating: $\displaystyle
    - M \leqslant f'\left( x \right) \leqslant M \Rightarrow \int_0^t { (- M)dx} \leqslant \int_0^t {f'\left( x \right)dx} \leqslant \int_0^t {Mdx}
    $ whenever $\displaystyle
    0 \leqslant t \leqslant 1
    $

    By the FTC: $\displaystyle
    \int_0^t {f'\left( x \right)dx} = f\left( t \right) - f\left( 0 \right)
    $

    Thus: $\displaystyle
    \left( { - M} \right) \cdot t \leqslant f\left( t \right) - f\left( 0 \right) \leqslant M \cdot t

    $

    Integrating again: $\displaystyle
    \left( { - M} \right) \cdot \int_0^1 {t \cdot dt} \leqslant \int_0^1 {f\left( t \right) \cdot dt} - \int_0^1 {f\left( 0 \right) \cdot dt} \leqslant M \cdot \int_0^1 {t \cdot dt}
    $

    Therefore: $\displaystyle
    \left( {\tfrac{{ - M}}
    {2}} \right) \leqslant \int_0^1 {f\left( t \right) \cdot dt} - f\left( 0 \right) \leqslant \left( {\tfrac{M}
    {2}} \right)
    $

    Since $\displaystyle f(0)=0$ we get the desired result

    All the other parts are similar, try to do them and then ask if you still have doubts.

    1

    how can u get the second ques?
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