1. We are given $g(x)=\int_0^x f(t)\,dt$ and need to find the largest possible interval in which $g(3)$ must lie.

2. Split the integral at $x=1$: $g(3)=\int_0^3 f(t)\,dt=\int_0^1 f(t)\,dt+\int_1^3 f(t)\,dt.$

3. Use the bounds on $f$ over each interval.

On $[0,1]$: $\frac13\le f(t)\le 1.$ Since the interval length is $1$, $\int_0^1 \frac13\,dt \le \int_0^1 f(t)\,dt \le \int_0^1 1\,dt,$ so $\frac13 \le \int_0^1 f(t)\,dt \le 1.$

4. On $(1,3]$: $0\le f(t)\le \frac12.$ The interval length is $2$, so $\int_1^3 0\,dt \le \int_1^3 f(t)\,dt \le \int_1^3 \frac12\,dt,$ thus $0\le \int_1^3 f(t)\,dt \le 1.$

5. Add the two bounds: $\frac13+0 \le g(3) \le 1+1,$ which gives $\frac13\le g(3)\le 2.$

6. So the largest possible interval containing all possible values of $g(3)$ is $\left[\frac13,2\right].$

7. Check options:

• A and B are negative intervals, impossible since $f(t)\ge 0$ everywhere.
• C: $[1,3]$ is too restrictive and also upper bound $3$ is impossible.
• D: $\left[\frac13,2\right]$ matches exactly.

Therefore, the correct option is D.