show its integral less than half M

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April 8th 2008, 09:07 AM
szpengchao

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?
April 8th 2008, 10:18 AM
PaulRS

Quote:

Originally Posted by szpengchao

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?

$ \left| {f'\left( x \right)} \right| \leqslant M \Leftrightarrow - M \leqslant f'\left( x \right) \leqslant M $

Integrating: $ - 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 $ 0 \leqslant t \leqslant 1 $

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

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

Integrating again: $ \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: $ \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 $f(0)=0$ we get the desired result

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

1:):)
April 8th 2008, 10:33 AM
szpengchao

how can u get the second ques?

Quote:

Originally Posted by PaulRS

$ \left| {f'\left( x \right)} \right| \leqslant M \Leftrightarrow - M \leqslant f'\left( x \right) \leqslant M $

Integrating: $ - 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 $ 0 \leqslant t \leqslant 1 $

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

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

Integrating again: $ \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: $ \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 $f(0)=0$ 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?