Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having some trouble with the following problem. I've listed what I've used so far.

(Edit: Made a mistake with the second integral.)

The problem is as follows:

If a function $f: [0,1] \rightarrow \mathbb R$ is continuous and $\int_0^x f = \int_x^1 f, \forall x \in [0,1]$, show that $f(x) = 0, \forall x \in [0,1]$.

I would assume a proof by contradiction. So assume that $\exists x \in [0,1]$ such that $f(x) \neq 0$.

By the FTOC, since $\int_0^x f = \int_x^1 f$, it follows that if $F'(x) = f(x), \forall x \in [0,1]$, then $F(x)-F(0) = F(1) - F(x)$, or equivalently:

$F(x) = \dfrac{F(1)+F(0)}{2}$.

I'm wondering whether the given approach is correct or whether I should prove the question in a different approach.

Any feedback would be appreciated!

share|improve this question

1 Answer 1

You have that $$\int_0^x f=\int_x^1 f$$ for each $x\in[0,1]$. By FTC, upon differentiation, you get that $$f(x)=-f(x)$$ for each $x\in [0,1]$. That is $2f(x)=0$ or $f(x)=0$ over $[0,1]$.

share|improve this answer
    
Ah, sorry Peter. I've made a mistake to the question! There should be an edit. –  Julian Park Apr 7 '13 at 1:15
    
@JulianPark OK... no worries. –  Pedro Tamaroff Apr 7 '13 at 1:15
    
@JulianPark See the edit. –  Pedro Tamaroff Apr 7 '13 at 1:18
    
Thank you Peter! –  Julian Park Apr 7 '13 at 1:21
    
@JulianPark You can accept the answer if you feel like it. –  Pedro Tamaroff Apr 7 '13 at 17:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.