integration of continuous function defined on [0,1]

Let $$f:[0,1]\to{\mathbb{R}}$$ be continuous such that $$f(t)\ge0$$ for all $$t \in [0,1]$$. Define $$g(x)=\int_{0}^{x}f(t)\,dt$$. Then

1.$$g$$ is monotone and bounded

2.$$g$$ is monotone but not bounded

3.$$g$$ is bounded but not monotone

4.$$g$$ is neither monotone nor bounded

$$f$$ is continuous on compact set hence image must be compact say $$[a,b]$$ where $$a,b\ge0$$

$$g(x)=0$$ for $$x\in[0,a]$$

$$g(x)\ge0$$ for $$x\in[a,b]$$

$$g(x)=\int_{a}^{b} f(t)\,dt$$ for $$x\ge{b}$$ that is constant

hence first option correct

please correct me if i am wrong

Your answer is almost correct. A correct justification of the boundeness part would be: $(\forall x\in[0,1]):g(x)=\int_0^xf(t)\,dt\leqslant\int_0^1f(t)\,dt$. Hence, $g$ is bounded above. Of course, it is bounded below, since $(\forall x\in[0,1]):g(x)\geqslant0$.
Let $X,Y \in [0,1] : X<Y$
$$g (Y)-g (X)=\int_X^Y f (t)dt\geq 0$$
then $g$ is increasing at $[0,1]$. or by derivative. $f$ is continuous at $[0,1] \implies g$ is differentiable at $[0,1]$ and $$g'(x)=f (x)\ge 0.$$
on the other hand, $f$ continuous at the compact $[0,1] \implies f$ is bounded. $\implies \exists M \ge 0\;\;:$ $$\forall t\in [0,1]\;\;0\le f (t) \le M$$ which yields by integration to $$0\le g (x)\le M (x-0) \le M$$ thus $g$ is bounded.