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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

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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 $(\forall x\in[0,1]):g(x)\geqslant0$.

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hint

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.

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