Let $$g(x) = \int_0^x f(t)dt$$, where $$f$$ is continuous function in $$[0, 3]$$ such that $$\frac{1}{3} \le f(t) \le 1$$ for all $$t \in [0, 1]$$ and $$0 \le f(t) \le \frac{1}{2}$$ for all $$t \in (1, 3]$$.
The largest possible interval in which $$g(3)$$ lies is :

We have $$g(3) = \int_0^3 f(t)\,dt = \int_0^1 f(t)\,dt + \int_1^3 f(t)\,dt$$.

For $$t \in [0,1]$$, we know $$\frac{1}{3} \le f(t) \le 1$$, so $$\int_0^1 \frac{1}{3}\,dt \le \int_0^1 f(t)\,dt \le \int_0^1 1\,dt$$, giving $$\frac{1}{3} \le \int_0^1 f(t)\,dt \le 1$$.

For $$t \in (1,3]$$, we know $$0 \le f(t) \le \frac{1}{2}$$, so $$\int_1^3 0\,dt \le \int_1^3 f(t)\,dt \le \int_1^3 \frac{1}{2}\,dt$$, giving $$0 \le \int_1^3 f(t)\,dt \le 1$$.

Adding the bounds: $$\frac{1}{3} + 0 \le g(3) \le 1 + 1$$, so $$\frac{1}{3} \le g(3) \le 2$$. The largest possible interval in which $$g(3)$$ lies is $$\left[\frac{1}{3}, 2\right]$$.