Prove that if integral of a squared function is zero, then function is zero function I almost got this proof done but I can't seem to justify a little step. It goes:

Let $f$ be a real-valued, continuous function on $[a,b]$. Prove that if $$\int_a^b [f(x)]²\ dx = 0$$ then $$f(x)=0 \,\,\,\,\,\,\,\,\forall x \in [a,b]$$

I start by defining $$F(x) = \int_a^x [f(t)]²\ dt $$ Since $f$ is continuous, then $F'(x)=[f(x)]²\ge0 \;\;\;\;\; \forall x \in [a,b]$. Thus $F$ is increasing in $[a,b]$.
It's clear that $F(a)=0$ and by hypothesis $F(b)=0$. If I could justify why this means that $F$ must be a constant function in $[a,b]$ then my proof would be completed, since that would mean $F'(x)=0\;\;\forall x\in[a,b]$ and therefore $f(x)=0\;\;\forall x\in[a,b]$. 
Could anyone please tell me if there's a theorem or anything that would let me justify the key step???

To complete the proof along the lines you started: $F$ is increasing (non-decreasing) on the entire interval. $F(a) = F(b)$. For any $x$ between $a$ and $b$, $F(a) \le F(x) \le F(b) = F(a)$. So the inequalities must be equalities.
