- #1
steven187
- 176
- 0
hello all
well iv been workin on this problem but no matter how many ideas cross my mind i can't prove it, iv tried playing around with the properties of continuous 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 don't 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
well iv been workin on this problem but no matter how many ideas cross my mind i can't prove it, iv tried playing around with the properties of continuous 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 don't 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