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Let $f:[0,1]\to\mathbb R$ be continuous such that $f(t)\geq 0$ for all $t$ in $[0,1]$. What can be said about $g(x):=\int_0^x f(t)\,dt$?

Let $f:[0,1] \to\mathbb{R}$ be continuous such that $f(t) ≥0$ for all t in $[0, 1]$. Define $g(x) = \int_0^xf(t) \, dt$ then which is true?
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

I think either 1 or 2 is true as I get it is monotone but not sure about boundedness

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marked as duplicate by Cameron Buie, David Mitra, Davide Giraudo, Martin Sleziak, QiL Dec 17 '12 at 16:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Hint: $f$ is bounded on $[0,1]$. – David Mitra Dec 17 '12 at 15:12

Since $f$ is continuous and defined in a closed interval, $f$ is bounded by say $m,M$ below and above, respectively. Then $$mx= \int_0^x m \, dt\le \int_0^x f(t) \, dt\le \int_0^x M \, dt=Mx$$ What does this say about $g$ in $[0,1]$?

An alternate approach without assuming continuity of $f$ (and thus differentiability of $g$) is the following:

$g$ is Lipschitz continuous and defined in a closed interval, thus it is bounded. In addition since $f(t)\ge 0$ for $t\in [0,1]$, if $0\le x_1<x_2\le 1$, $$g(x_2)-g(x_1)=\int_{0}^{x_2}f(t)\, dt-\int_{0}^{x_1}f(t)\, dt=\int_{x_1}^{x_2}f(t)\, dt\ge 0$$

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