Inverse of Definite Integrals? 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.
 A: If $ f $ is integrable at $ [a,b]$, then the function $ g $ defined by
$$g(x)=\int f(x)dx$$
is not injective so $ g^{-1}(x)$ doesn't exist.
For example, if $ f(x)=\cos(x)$ then
$$g(x)=\sin(x)\text{ and } g(x)=\sin(x)+1$$
In the second case, if
$$g(x)=\int_a^bf(x)dx $$
then $ g $ is constante and cannot be bijective.
we can't speak about $ g^{-1}(x)$.
A: You have two very different kinds of "inverse" in the question.
When someone says that $g^{-1}(x)$ is the inverse of a function $g(x),$
they mean that if you take some arbitrary number $x$
in the domain of the function $g,$
and use this value of $x$ as input to the function $g$,
and the resulting output is some number $y,$
and you then take this number $y$ and use it as input for the function $g^{-1},$
the result then will be the original number $x$ that you started with.
In short:
If $g(x) = y$ then $g^{-1}(y) = x.$
That's what an inverse of a function is.
When people say the inverse of an integral is the derivative, they mean that you start with some function $f$ (not a number!),
you use indefinite integration to get another function $g$
-- technically, actually, you get a whole family of functions,
which is why in many books you will see each indefinite integral written in the form
$$ \int f(x)\,\mathrm dx = g(x) + C $$
where $C$ represents an arbitrary constant.
Then you can take any example of the the "integral" $g(x) + C$ for some particular constant $C$ -- for example we can say $C=0$ and just use the function $g$ --
and apply differentiation to that function,
and you will get back the original function $f.$
In short:
If $\ \int f(x)\,\mathrm dx = g(x) + C\ $
then $\ \dfrac{\mathrm d(g(x))}{\mathrm dx} = f(x).$
That's all people mean when they say "the derivative is the inverse of the integral". They are not saying anything about $g^{-1}(x)$.
Personally, I would not even say that "the derivative" is the inverse of "the integral"; I would say differentiation is the inverse of (indefinite) integration.
One reason I would not say that "the derivative" and
"the integral" are inverses is that those are not the best words:
in most contexts, the words "the derivative" and "the integral"
refer to functions that we get from other functions,
whereas the two things that we are trying to say are inverses are the
ways we get the functions from each other.
It's very rare that "the derivative" actually means differentiation
or that "the integral" actually means integration.
(And I can't think of any time I have ever heard someone refer to
integration as "an integral".)
A second reason is that using words in this way gives people wrong ideas and leads them into potentially endless confusion.
So far I haven't even gotten to the actual question about the definite integral, but the thing is that if you start with a bad premise you don't get a good question.
Regarding the definite integral, for any two real numbers $a$ and $b$ with
$a < b$ and for any function $f$ that is integrable on the interval $[a,b],$
the definite integral
$$ \int_a^b f(x)\,\mathrm dx $$
is just a number. It's not a function, though you can use it like any other number to define a constant function.
There were three things -- $a,$ $b,$ and the function $f$ -- that we used as inputs (of some kind) to that definite integral. The result of the definite integral could be said to depend on any one of the three things, or all three of them.
But you cannot invert the process of definite integration to get back any of those three things, because all you have at the end of definite integration is a single number that could have been the result of unimaginably many very different definite integrals.
So there is no inverse of definite integration.
