1. ## continuous function, integral

Suppose that $f: [a,b] \rightarrow \mathbb{R}$ is continuous and $f(x) \geq 0$ for all $x \in [a,b]$. Prove that if $\int^b_a f(x)dx=0$, then $f(x)=0$ for all $x \in [a,b]$.
Attempt
I had attempted to do this problem by contradiction, except I did not understand how to finish the problem. I would appreciate a few helpful hints on this one.

2. ## Fundamental Proof

Originally Posted by selenne431
Suppose that $f: [a,b] \rightarrow \mathbb{R}$ is continuous and $f(x) \geq 0$ for all $x \in [a,b]$. Prove that if $\int^b_a f(x)dx=0$, then $f(x)=0$ for all $x \in [a,b]$.
From the proof in the fundamental theorem of calculus:

$F(b) - F(a) = \lim_{\| \Delta \| \to 0} \sum_{i=1}^n \,[f(c_i)(\Delta x_i)]\$

$F(b) - F(a) = \int_{a}^{b} f(x)\,dx\$

With $f(c_i)\geq 0$ from given $f(x) \geq 0$ for all $x \in [a,b]$

And the definite integral evaluated as $\int^b_a f(x)dx=0$ from given, then
$\lim_{\| \Delta \| \to 0} \sum_{i=1}^n \,[f(c_i)(\Delta x_i)]=0\$

with $f(c_i)$ confined to be non negative, the summation can only be zero if $f(x)=0$ for all $x \in [a,b]$.