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: 
To show not a function:

One idea could be to show that the line $x = 0$ will intersect the rotated graph of $y = x^2$. To do this you could notice that this line would have had equation $y = \tan(\frac{\pi}{2}-\theta) \cdot x = \cot(\theta) \cdot x$ prior to the rotation. We can solve $\cot(\theta) \cdot x = x^2$. We get that $x = 0$ and $x = \cot(\theta)$ give two intersection points for the parabola and the line prior to the rotation.

In the "bad" case that $\cot(\theta) = 0$ we know that the graph of the rotated parabola is symmetric about the $x$-axis.
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. $\square$

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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. $\square$

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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.