How to prove that
If $\;f,g:\big[0,+\infty\big)\to\big[0,+\infty\big)\;$ are continuous functions and
$\;f(x)=O\big(g(x)\big)\;$ as $\;x\to 0;,\;$ then
$\;F(x)=O\big(G(x)\big)\;$ as $\;x\to 0;,$
where $\;\displaystyle F(x)=\int_0^x f(t)dt\;$ and $\;\displaystyle G(x)=\int_0^x g(t)dt\;.$ can I assume that $f(x) = O(\mid x \mid ^a)$ and $g(x) = O(\mid x \mid ^b)$?
It is not true, there are a counterexample :
$f(x)=\begin{cases} \sin x\quad\text{ if }x\in\big[0,\pi\big]\\ 0\qquad\;\text{ if }x\in\big]\pi,+\infty\big[ \end{cases}\quad,$
$g(x)=0\quad\text{ for all }x\in\big[0,+\infty\big[\;.$
It results that
$f,g:\big[0,+\infty\big[\to\big[0,+\infty\big[\;$ are continuous functions ,
$f(x)=O\big(g(x)\big)\;$ as $\;x\to+\infty\;,$
indeed there exist $\;M=1>0\;$ and $\;x_0=\pi\;$ such that
$0\leqslant f(x)\leqslant Mg(x)\;$ for all $\;x\geqslant x_0\;.$
But
$F(x)\ne O\big(G(x)\big)\;$ as $\;x\to+\infty\;,$
indeed
$F(x)=\displaystyle\int_0^x f(t)dt=\begin{cases} 1-\cos x\quad\text{ if }x\in\big[0,\pi\big]\\ 2\qquad\qquad\text{ if }x\in\big]\pi,+\infty\big[ \end{cases}\;,$
$G(x)=\displaystyle\int_0^x g(t)dt= 0\quad\text{ for all }x\in\big[0,+\infty\big[\;,$
consequently there do not exist any $\;M>0\;$ and any $x_0\geqslant0\;$ such that
$0\leqslant F(x)\leqslant MG(x)\;$ for all $\;x\geqslant x_0\;.$
Addendum :
If we want that the property of the original poster is true, it is necessary to add another hypothesis.
An additional hypothesis could be the following one :
there exists $\;x^*\geqslant0\;$ such that $\;g(x^*)>0\;.$
Theorem :
If $\;f,g:\big[0,+\infty\big[\to\big[0,+\infty\big[\;$ are continuous functions ,
there exists $\;x^*\geqslant0\;$ such that $\;g(x^*)>0\;$ and
$f(x)=O\big(g(x)\big)\;$ as $\;x\to+\infty\;,\;$ then
$\;F(x)=O\big(G(x)\big)\;$ as $\;x\to+\infty\;,$
where $\;\displaystyle F(x)=\int_0^x f(t)dt\;$ and $\;\displaystyle G(x)=\int_0^x g(t)dt\;.$
Proof :
From hypothesis we know that
$f(x)=O\big(g(x)\big)\;$ as $\;x\to+\infty\;,$
so there exist $\;M>0\;$ and $\;x_0\geqslant0\;$ such that
$0\leqslant f(x)\leqslant Mg(x)\;$ for all $\;x\geqslant x_0\;.\quad\color{blue}{(*)}$
Since $\;g(x)\;$ is a non-negative continuous function and there exists $\;x^*\geqslant0\;$ such that $\;g(x^*)>0\;,\;$ then it follows that
$\displaystyle\int_0^{x^*+1}g(t)dt>0\;.$
By letting $\;x_1=\max\big\{x_0,\;x^*+1\big\}>0\;,\;$ we get that
$\displaystyle\int_0^{x_1}g(t)dt\geqslant\int_0^{x^*+1}g(t)dt>0\;.$
Moreover, by letting $\;N=\max\left\{M,\;\dfrac{\int_0^{x_1}f(t)dt}{\int_0^{x_1}g(t)dt}\right\},$
we get that :
$\displaystyle\int_0^{x_1}f(t)dt\leqslant N\!\int_0^{x_1}g(t)dt\;.\quad\color{blue}{(**)}$
From $\;(*)\;$ it follows that
$\displaystyle\int_{x_1}^x\!\!\!f(t)dt\leqslant M\!\!\int_{x_1}^x\!\!\!g(t)dt\leqslant N\!\!\int_{x_1}^x\!\!\!g(t)dt\;\;$ for all $\;x\geqslant x_1\;,$
$\displaystyle\int_{x_1}^x\!f(t)dt\leqslant N\!\!\int_{x_1}^x\!g(t)dt\;\;$ for all $\;x\geqslant x_1\;.\quad\color{blue}{(*\!*\!*)}$
By adding the inequalities $\;(**)\;$ and $\;(*\!*\!*)\;,\;$ we get that
$\displaystyle\int_0^x\!f(t)dt\leqslant N\!\!\int_0^x\!g(t)dt\;\;$ for all $\;x\geqslant x_1\;.$
Hence, there exist $\;N>0\;$ and $\;x_1>0\;$ such that
$0\leqslant F(x)\leqslant NG(x)\;$ for all $\;x\geqslant x_1\;$
and it means that
$F(x)=O\big(G(x)\big)\;$ as $\;x\to+\infty\;.$
