Prove that $\int_{I}f=0 \iff$ the function $f\colon I\to \Bbb R$ is identically $0$. Let $I$ be a generalized rectangle in $\Bbb R^n$ 
Suppose that the function $f\colon I\to \Bbb R$ is continuous. Assume that $f(x)\ge 0$, $\forall x \in I$
Prove that $\int_{I}f=0 \iff$ the function $f\colon I\to \Bbb R$ is identically $0$. 

My idea is that
For $(\impliedby)$
Since $f\colon I\to \Bbb R$ is identically zero, $$f(I)=0$$
Then $$\int_{I}f=\int 0=0$$
For $(\implies)$,
Since $f$ is continuous, the function is integrable. 
i.e $\int _{I} f $ exists. 

I need the show that $\int f=0$ but how? 
Hopefully, other solution is true. Please check this. And how to continue this? Thank you:) 

One direction is obvious. If $f$ is identically $0$, the integral is zero. Now suppose $f$ is continuous but not identically zero. Then there exists a point in $I$ such that $f(\xi)>0$. By continuity there exists a neighborhood of $\xi$ where $f(x)>0$ whenever $x\in(\xi-\delta,\xi+\delta)$. Can you take it from here?
Since $f\geqslant 0$, it follows that $$\int_I f \geq \int_{(\xi-\delta,\xi+\delta)} f>0$$
Thus, we have proven the contrapositive.
