$f:[a,b] \to R$ 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 $f(x_0)>0$ for some $x_0\in(a,b)$. Then there exists $\delta>0$ such that, for $|x-x_0|<\delta$, $f(x)>k>0$ (take $k=f(x_0)/2$, for instance), with $(x_0-\delta,x_0+\delta)\subseteq(a,b)$.
Now build a function $g$ by decreeing that
$$
g(x)=\begin{cases}
0 & \text{if $a\le x< x_0-\delta$}\\[3px]
? & \text{if $x_0-\delta\le x<x_0-\delta/2$}\\[3px]
1 & \text{if $x_0-\delta/2\le x\le x_0+\delta/2$}\\[3px]
? & \text{if $x_0+\delta/2<x\le x_0+\delta$}\\[3px]
0 & \text{if $x_0+\delta<x\le b$}
\end{cases}
$$
where $?$ stands for the values of the segment that makes the function continue in the two intervals (you can easily compute the formula).
Then $f(x)g(x)=0$ outside of $(x_0-\delta,x_0+\delta)$, non negative in this interval and strictly positive on $[x_0-\delta/2,x_0+\delta/2]$, so
$$
\int_{a}^b f(x)g(x)\,dx>0
$$
