If $f(x)=a+bx+o(x)$ why $\int f =ax+bx^2/2+o(x^2)$?

Question

Let $f(x)=a+bx+o(x)$. Why $$\int_0^xf(s)ds=ax+\frac{bx^2}{2}+o(x^2)\ \ ?$$

My problem is rather : why $$\int o(x)dx=o(x^2) \ \ ?$$

I know that since $f(x)=a+bx+o(x)$, there is $\varepsilon (x)$ s.t. $$f(x)=a+bx+x\varepsilon (x),$$ with $\varepsilon (x)\to 0$ whe $x\to 0$. In particular, $\varepsilon (x)$ is continuous on a neighborhood of $0$ (since $$\varepsilon (x)=\frac{f(x)-a-bx}{x}$$ and $\lim_{x\to 0}\varepsilon (x)=0$. Therefore it's integrable. But now, why $$\int_0^x t\varepsilon (t)dt=t^2\tau(x)$$ where $\tau(x)\to 0$ ?

Answer 1

Method 1

Set $F(x)=\int_0^x f(t)\,\mathrm d t$. Then $$F(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+o(x^2)$$ $$=f(0)x+\frac{f'(0)}{2}x^2+o(x^2)=ax+\frac{b}{2}x^2+o(x^2).$$

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Method 2 (Using your idea)

Since $\varepsilon (x)\to 0$ when $x\to 0$, there is $M>0$ and $\delta >0$ s.t. $|\varepsilon (x)|\leq M$. Therefore, if $|x|<\delta $, $$\left|\int_0^x t\varepsilon (t)\,\mathrm d t\right|\leq Mx^2\cdot |x|=o(x^2).$$