A formal proof required using real analysis If $\int_0^1f^{2n}(x)dx=0$, prove that $f(x)=0$, where $f$ is a real valued continuous function on [0,1]?
It is obvious, since $f^{2n}(x) \geq 0$, the only way this is possible is when $f(x)=0$.
I am looking for any other formal way of writing this proof i.e. using concepts from real analysis. 
Any hints will be appreciated.

Assume there exists $c\in[0,1]$ such that $f^{2n}(x)>0$, then by definition of continuity there exists $0<\delta< \min(c,1-c)$ such that $$|x-c|<\delta\implies |f^{2n}(x)-f^{2n}(c)|<\frac{1}{2} f^{2n}(c)$$
Especially, if $|x-c|<\delta$ then $f^{2n}(x)>\frac{1}{2} f^{2n}(c)$. Therefore
$$\int_0^1 f^{2n}(x) dx =\int_0^{c-\delta} f^{2n}(x) dx+\int_{c-\delta}^{c+\delta} f^{2n}(x) dx+\int_{c+\delta}^1 f^{2n}(x) dx\\
\ge 0+2\delta \int_{c-\delta}^{c+\delta} \frac{f^{2n}(c)}{2} dx+0 >0 $$
