Results 1 to 3 of 3

Math Help - show its integral less than half M

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    321

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    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?
    <br />
\left| {f'\left( x \right)} \right| \leqslant M \Leftrightarrow  - M \leqslant f'\left( x \right) \leqslant M<br />

    Integrating: <br />
 - 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} <br />
whenever <br />
0 \leqslant t \leqslant 1<br />

    By the FTC: <br />
\int_0^t {f'\left( x \right)dx}  = f\left( t \right) - f\left( 0 \right)<br />

    Thus: <br />
\left( { - M} \right) \cdot t \leqslant f\left( t \right) - f\left( 0 \right) \leqslant M \cdot t<br /> <br />

    Integrating again: <br />
\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} <br />

    Therefore: <br />
\left( {\tfrac{{ - M}}<br />
{2}} \right) \leqslant \int_0^1 {f\left( t \right) \cdot dt}  - f\left( 0 \right) \leqslant \left( {\tfrac{M}<br />
{2}} \right)<br />

    Since 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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
    321

    how can u get the second ques?

    Quote Originally Posted by PaulRS View Post
    <br />
\left| {f'\left( x \right)} \right| \leqslant M \Leftrightarrow - M \leqslant f'\left( x \right) \leqslant M<br />

    Integrating: <br />
- 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} <br />
whenever <br />
0 \leqslant t \leqslant 1<br />

    By the FTC: <br />
\int_0^t {f'\left( x \right)dx} = f\left( t \right) - f\left( 0 \right)<br />

    Thus: <br />
\left( { - M} \right) \cdot t \leqslant f\left( t \right) - f\left( 0 \right) \leqslant M \cdot t<br /> <br />

    Integrating again: <br />
\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} <br />

    Therefore: <br />
\left( {\tfrac{{ - M}}<br />
{2}} \right) \leqslant \int_0^1 {f\left( t \right) \cdot dt} - f\left( 0 \right) \leqslant \left( {\tfrac{M}<br />
{2}} \right)<br />

    Since 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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integral - show that
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 16th 2010, 03:56 PM
  2. Show that this integral is divergent
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 19th 2010, 02:21 AM
  3. Integral around half an ellipse
    Posted in the Calculus Forum
    Replies: 0
    Last Post: January 7th 2010, 10:08 AM
  4. Half Width Half Maximum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 27th 2008, 04:30 AM
  5. Replies: 1
    Last Post: November 20th 2006, 04:08 PM

Search Tags


/mathhelpforum @mathhelpforum