Is there always a general formula for the linear transformation that maps polynomials to one of their antiderivatives? In Friedberg's Linear Algebra this example of a linear transformation comes up: $T:P_2(R) \rightarrow P_3(R)$, such that $T(p)=q$, with $q(x)=\int_0^x{p}$, where $P_n(R)$ is the vector space of all polynomials over the reals of degree up to $n$.
At first my intuition suggested that, given any of a polynomial's antiderivatives, there should exist a linear transformation that maps that polynomial to that specific antiderivative, and that that would be accomplished by suitably varying the integral's lower bound in the preceding expression. However, consider the polynomial $f$ such that $f(x)=x$, and take its integral for an arbitrary lower bound $C$:
$$g_C(x)=\int_C^x f=\frac{x^2}{2}-\frac{C^2}{2}$$
In other words, $g_C$ is of the form $g_C(x)=\frac{x^2}{2}-D$, for $D\geq0$. So no matter what lower bound we choose, we'll never obtain an antiderivative for which $g_c(0)>0$, which seems quite surprising. Despite this, nothing stops us from defining a linear transformation $U$ such that $U(f)=h$, with $f(x)=x$ and $h(x)=\frac{x^2}{2}+1$, and analogously for the other basis vectors, whichever basis we might be using.
I wasn't able to, however, find a general expression written in terms of basic operations and integrals that could express what this linear transformation does, as can be done for $T$, which, again, seems to me quite surprising. Put informally, why should certain antiderivatives be "privileged" with respect to others in the realm of linear transformations and their expressions? Or am I just not seeing what this general expression should be? Is there one?
 A: You don't have to restrict yourself to real values of $C$! For complex values of $C$ we can consider a line integral over any path from $C$ to $x$ (the answer doesn't depend on the choice of path) and this produces $\frac{x^2}{2} - \frac{C^2}{2}$ where the value of $\frac{C^2}{2}$ can now be arbitrary.
Of course we now run into the issue that integrating $1$ no longer produces a real polynomial.
Now for a general discussion which hopefully should clear some things up. There's no need to restrict ourselves to a specific degree so let's consider the derivative as a linear transformation $D : P_{n+1}(\mathbb{R}) \to P_n(\mathbb{R})$. We'll say that an antiderivative transformation is any right inverse to $D$: that is, any linear map $A : P_n(\mathbb{R}) \to P_{n+1}(\mathbb{R})$ satisfying $DA = I$. For any constant $C$, the linear map
$$A_C : f(x) \mapsto \int_C^x f(t) \, dt$$
is an antiderivative transformation. However, in general there are other antiderivative transformations!
We can define an antiderivative transformation by selecting its value on each element of the standard basis $\{ x^k \}$ of polynomials. This value can be any antiderivative of $x^k$, so can have the form $\frac{x^{k+1}}{k+1} + C_k$ where $C_k$ is a constant that depends on $k$. For $A_C$ this dependence takes the form $C_k = - \frac{C^{k+1}}{k+1}$ which is a very specific functional form; in general we can just select $n+1$ arbitrary constants.

I wasn't able to, however, find a general expression written in terms of basic operations and integrals that could express what this linear transformation does

This can be done but you're not going to like it. The general antiderivative on polynomials takes the form $A(f) = \int_0^x f(t) \, dt + C(f)$ where $C : P_n(\mathbb{R}) \to \mathbb{R}$ is any linear functional (this is a nice exercise). Every linear functional on $P_n$ can be written as an inner product
$$C(f) = \int_0^1 f(t) g(t) \, dt$$
(or any other fixed interval you like) for some other polynomial $g$ of degree $\le n$. So the general antiderivative on $P_n$ is
$$A(f) = \int_0^x f(t) \, dt + \int_0^1 f(t) g(t) \, dt$$
for some polynomial $g$. Other representations of linear functionals on polynomials are also possible; for example it's a nice exercise to show that every linear functional on polynomials of a fixed degree is a linear combination of evaluation maps $f \mapsto f(c)$.
A: I don't know if this will be satisfying as an answer to the question of why certain antiderivatives are "privileged", as you put it, but I'm not sure if that question has a good answer, so here are my thoughts:

given any of a polynomial's antiderivatives, there should exist a linear transformation that maps that polynomial to that specific antiderivative

There is, and you can construct it be choosing a basis containing that polynomial, and then defining the linear transformation to send the origin polynomial to the specified antiderivative, and the other elements of the basis to other polynomials in some basis of $P_3(\mathbb{R})$.

and that that would be accomplished by suitably varying the integral's lower bound

You've shown that this doesn't work over $\mathbb{R}$, but in some sense you're right about your idea, but just working over the wrong field. If you took polynomials with coefficients in $\mathbb{C}$, then you wouldn't have the problem that you ran into here. In general, your expression will give you the antiderivative minus a polynomial in the lower bound. Over $\mathbb{R}$ some polynomials are not surjective functions to $\mathbb{R}$, as in your example, but over $\mathbb{C}$ they are all surjective. This is a consequence of the fact that $\mathbb{C}$ is algebraically closed.
Edit (you always realize the right way to say something once you've already said it): To summarize what I said above and connect it to what's "privileged" about certain antiderivatives, it's the fact that for any polynomial $f(x)$, we'll have $\int_C^x f\ dx = F(x) - F(C)$, where $F$ is the antiderivative of $f(x)$. But if $F(x)$ isn't surjective onto $\mathbb{R}$, you will only get the antiderivatives with constant in the image of $F$. This can happen over $\mathbb{R}$, but not over $\mathbb{C}$, as mentioned above.
A: A basic viewpoint is considering the space of polynomials over real numbers $\mathbb R[X]$ as the vector space spanned by the basis of monomials $(X^n)_{n\in \mathbb N}$. Now you can define the antiderivative $A: \mathbb R[X]\to \mathbb R[X]$ by its action on the basis:
$$
A(X^n) = \frac{X^{n+1}}{n+1},\quad n\ge 0.
$$
