Sign of $\int_0^x f(t)dt$ Is right that $\int_0^x f(t)\,dt>0$ if $x>0$ and $f(t)$ is positive and also if $x<0$ and $f$ is negative?
My doubt is in the second: but I have thought that since if $x<0$ then $\int_0^x f(t)dt=-\int_x^0 f(t)dt$ and if $f(t)$ si negative then $\int_x^0 f(t)dt<0$ and so $-\int_x^0 f(t)dt>0$. Is it right?
 A: Notice that, if $m \leq f$ on $[0,x]$, then
$$\int_0^xf(t) \ dt \geq mx.$$
If $f$ is positive ($m>0$) and $x>0$, then $mx>0$. If $f$ is negative ($m<0$) and $x<0$, then $mx>0$.
Edit: As pointed out in a comment, one can have $f$ intregable with $f>0$ on $[0,x]$, but $f \not\geq m$ for any $m>0$. An example of this might be
$$f(t) = \begin{cases} t & 0<t\leq x\\ 1 & t=0 \end{cases}.$$
In that case, as long as we have $f\geq m > 0$ on some small subinterval $I \subset [0,x]$ of length $\varepsilon>0$, then
$$\int_0^x f(t) \ dt \geq \int_I f(t) \ dt \geq m\varepsilon > 0.$$
A: The definition of integral states that
$$
\int_0^xf(t)dt=\lim_{n\to\infty}\frac{x}{n}\sum_{k=1}^{n} f(\frac{kx}{n}),
$$
for $x>0$ and $\int_0^xf(t)dt=-\int_x^0f(t)dt$. If $f(t)>0$ for $t>0$ we obtain $\int_0^xf(t)dt\ge 0$.
A: $$\int_0^xf(x)\,dx=-\int_x^0f(x)\,dx$$ answers the question.
A: okay so if we have $f(t)>0\forall t,\,x>0$ then we can say:
$$\int_0^xf(t)dt>0$$
however if we are to take:
$$\int_0^{-x}f(t)dt=-\int_{-x}^0f(t)dt<0$$
and this is due to the fundamental theorem of calculus (switching limits) and due to the fact $f(t)>0$ and so the integral is too. Hope this is what you were looking for :)
