$f:[a,b] \to R$ is continuous and $\int_a^b{f(x)g(x)dx}=0$ for every continuous function $g:[a,b]\to R$ $f:[a,b] \to R$ is continuous and $\int_a^b{f(x)g(x)dx}=0$ for every continuous function $g:[a,b]\to R$ with $g(a)=g(b)=0$. Must $f$ vanish identically?

Using integration by parts I got the form:
$\int_a^bg(x)f(x)-g'(x)F(x)=0$. Where $F'(x)=f(x)$.

Suppose that there exists $x_0$ that $f(x_0)=\varepsilon \neq 0$, we can assume that $\varepsilon>0$ without loosing of generality. $f$ is continuous, so there exists $\delta$ such that for $x_1 \in (x_0-\delta,x_0+\delta) \subset (a,b)$ we have $f(x_1) > \frac{\varepsilon}{2}$. Now you can find such a function $g$, that $g(x_2)>1$ for $x_2 \in (x_0-\frac{\delta}{2},x_0+\frac{\delta}{2})$  and $g(x) \geq 0$ (for example piecewise linear), then you have:
$$\int_{a}^{b}f(x)g(x)dx \geq \int_{x_0-\frac{\delta}{2}}^{x_0+\frac{\delta}{2}}f(x)g(x)dx \geq 1 \cdot 2\delta \cdot \frac{\varepsilon}{2}=\delta \varepsilon>0$$
