Proof concerning definite integral: $ \int^{b}_{a} f(x) \: dx = 0 \iff f = 0 $ Let $f \in C^0 ([a, b], \mathbb{R})$ and $f \geq 0$. Show that 
$$ \int^{b}_{a} f(x) \: dx = 0 \iff f = 0 $$
The $ \Leftarrow $ part I've already proved. Let us consider the $ \Rightarrow $ part:
\begin{equation*}
\begin{split}
\int^{b}_{a} f(x) \: dx = F(b) - F(a) & = 0 \\
F(b) & = F(a)
\end{split} 
\end{equation*}
Is this the right approach? How I can go on?
 A: I don't think your approach will be fruitful.
Hint: Suppose $f\not\equiv0$ and show that in that case $\int_a^bf\neq0$.
Details: More explicitly, if $f\not\equiv0$ then there exists $\xi\in[a,b]$ such that $f(\xi)=:\epsilon>0$. The continuity of $f$ implies that there exists a non-degenerate interval $I$ of $[a,b]$ containing $\xi$ on which, say, $f\geq\frac{\epsilon}{2}$. Then since $f$ is nonnegative, $\int_a^b f\geq\int_I f\geq\ell(I)\frac{\epsilon}{2}>0$.
A: It is trivial that if $f=0$ then $\int_{a}^b f(x)dx=0$.
Now suppose that $\int_a^b f(x)dx=0$. We need to prove that $f=0$, that means $f(x)=0$ for all $x\in [a,b]$. We use the contradiction to solve. In fact, assume that $\exists x_0\in [a,b]$ such that $f(x_0)>0$. Since $f$ is continuous at $x_0$, there exist a $\delta>0$ small enough such that $f(x)>\frac{f(x_0)}{2}$ for all $x\in (x_0-\delta, x_0+\delta)$, (since
$$\varepsilon  = \frac{{f\left( {{x_0}} \right)}}{2} > 0,\,\,\exists \delta  > 0:\left| {x - {x_0}} \right| < \delta  \Rightarrow \left| {f\left( x \right) - f\left( {{x_0}} \right)} \right| < \frac{{f\left( {{x_0}} \right)}}{2}.)$$
Hence
$$\int_a^b {f\left( x \right)dx}  \geqslant \int_{{x_0} - \delta }^{{x_0} + \delta } {f\left( x \right)dx}  \geqslant \int_{{x_0} - \delta }^{{x_0} + \delta } {\frac{{f\left( {{x_0}} \right)}}{2}dx}  = \frac{{f\left( {{x_0}} \right)}}{2}\int_{{x_0} - \delta }^{{x_0} + \delta } {dx}  = \delta f\left( {{x_0}} \right) > 0.$$
This is a contradiction. So $f(x)=0$ for all $x\in [a,b]$.
A: Another answer (perhaps more linked to your tentative). Put $\displaystyle F(x)=\int_a^x f(t)dt$. As $f$ is continuous, the derivative of $F$ exists and is equal to $f$. As $f$ is positive, $F$ is increasing. Hence for $a\leq x\leq b$, we get $0=F(a)\leq F(x)\leq F(b)=0$. Hence $F=0$, and $f=F^{\prime}$ is $0$
