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