Question: We know that the inverse of an integral is the derivative, but what happens if the integral is definite (meaning: it has bounds/limits from-to)?

For example we have the function $g(x) = \int f(x) dx$, then to find $g^{-1}(x)$ we take the inverse of integral which is derivative and about this part im unsure, we probably take the inverse of $f(x)$ i guess (meanwhile $f^{-1}(x)$, also correct me if im wrong about this part)... In total, we have $g^{-1}(x) = \frac{d(f^{-1}(x))}{dx}$, or if im wrong, then it is $g^{-1}(x) = \frac{d(f(x))}{dx}$

But what if $g(x) = \int\_a^b f(x) dx$, and then what is $g^{-1}(x)$?

If there is a missing information in my question that is necessary to be put, just tell me so i add it.

Answer: 
The graph of $f$, rotated by $\theta$, would be the graph of a function if, and only if, it did not intersect any vertical line twice (the so-called "vertical line test").

A vertical line intersecting the graph of $f$, rotated by $\theta$, twice is the logically equivalent to a vertical line, rotated by $-\theta$, intersecting the graph of $f$ twice.

Let $m$ be the slope of a vertical line rotated by $-\theta$. We can take the slope of this line since $\theta$ is not a multiple of $\pi$ and the rotated line is therefore not vertical. Then consider the line $y = mx + 1$. I claim that this line intersects the given parabola twice.

Indeed, to find the x coordinates of an intersection, we must solve $x^2 = mx + 1$. That is, $x^2 - mx - 1 = 0$. Using the quadratic formula, we have $x = \frac{m \pm \sqrt{m^2 + 4}}{2}$. There are clearly 2 distinct solutions.
To show the equivalent notion that a non-vertical line meets an "upright" parabola in two points, consider the polar equation of such a parabola with its focus at the origin and with semi-latus rectum $p$:
$$r = \frac{p}{1-\sin\theta}$$
For $|\sin\theta|\neq1$ (that is, for non-vertical directions), angles $\theta$ and $\theta+\pi$ yield (finite) positive values for $r$, corresponding to points on the parabola in "opposite directions" from the focus. $\square$

---

Alternatively, but less concisely, the parabola rotated by angle $\phi$ has polar equation
$$r = \frac{p}{1-\sin(\theta-\phi)}$$
Points in the vertical directions correspond to $\theta=\pm\pi/2$, so we have
$$r = \frac{p}{1\pm\cos\phi}$$
For $|\cos\phi|\neq 1$, we get the two values of $r$, and the two points on the parabola on a vertical line through the the focus. $\square$

---

The switch to the polar context is a little "advanced", but the dead-simple way it identifies the points of intersection is tough to beat.