= Solution
Fix $t_0\geq0$. On a compact parameter <interval> containing all nonnegative $t$ sufficiently close to $t_0$, the corresponding rectangle with $x\in[0,1]$ is <compact>. The <continuous function> $f$ is therefore bounded there by some $M<\infty$, by the <extreme value theorem>.
For any sequence $t_n\geq0$ tending to $t_0$, all sufficiently late $t_n$ lie in this rectangle's parameter range. For each $x\in[0,1]$, <continuity> gives
$$
f(t_n,x)\longrightarrow f(t_0,x),\qquad |f(t_n,x)|\leq M.
$$
The constant <function> $M$ is integrable on $[0,1]$. The <dominated convergence theorem> therefore gives
$$
\boxed{F(t_n)=\int_0^1f(t_n,x)\,dx\longrightarrow\int_0^1f(t_0,x)\,dx=F(t_0).}
$$
The sequential criterion for <continuity> on a real <interval> proves \b[$F$ is continuous on $[0,\infty)$]. At zero this is right-continuity. This is <continuity of parameter integrals on compact rectangles>; the bound is local in the parameter, so $f$ need not be bounded on the entire unbounded strip.
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