hello all(adsbygoogle = window.adsbygoogle || []).push({});

well iv been workin on this problem but no matter how many ideas cross my mind i cant prove it, iv tried playing around with the properties of continous functions on closed intervals but didnt get me anywhere

let f be a contiuous function on [a,b] which is non-negative, suppose that

[tex]\int_{a}^{b}f(x)dx=0[/tex]

show that

[tex]f(x)=0 \ \forall x\epsilon [a,b][/tex]

this is probably the best of my proofs but i dont think its correct because i think i have to prove contradiction for all functions that are more than zero, please help any advice would be helpful

thank you

lets assume that f(x)>0 let f(x)=c>0 by FTC

[tex]\int_{a}^{b}f(x)dx=F(b)-F(a)=0[/tex]

[tex]F(b)=F(a)[/tex]

since f(x)>0 then F(x)=cx where c>=0

ca=cb

0=cb-ca

0=c(b-a) since x is an element of [a,b] for all a,b an element of R then c=0

therefore a contradiction and f(x)=0

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# To Prove

Loading...

Similar Threads - Prove | Date |
---|---|

I A problematic limit to prove | Jan 26, 2018 |

I Proving equivalence between statements about a sequence | Feb 12, 2017 |

I Prove that ∫f(x)δ(x)dx=f(0) | Jan 22, 2017 |

I Prove ln(x) <= x-1 for positive x | Jan 15, 2017 |

B How does the delta ε definition prove derivatives? | Aug 9, 2016 |

**Physics Forums - The Fusion of Science and Community**