Can $T:P_3\rightarrow P_4$ be a function of two variables? I'm given the following question:

Denote by $P_n$ the set of polynomials in the variable $x$ of degree at most $n$. The usual basis for $P_n$ is given by $\left\{1,x,x^2,\ldots,x^n\right\}$.
Define the linear transformation $T:P_3\rightarrow P_4$ by $T(p(x))=3p'(x)+\int_0^xp(t)dt$. Find the image of $T$.


Why is there $\boldsymbol{\int_0^xp(t)dt}$ in $T$? Neither basis is expressed in terms of $t$. Why can't we just use the same variable $x$?
My professor is a smart guy, and it's been awhile since we saw this question; could this be a typo? Does it make any sense?

We can re-write the formula for $T$ as
\begin{align*}
T(a+bx+cx^2+dx^3)
&= \frac{d}{dx}(a+bx+cx^2+dx^3)+\int_0^x(a+bt+ct^2+dt^3)\,dt \\
&= b+2\,cx+3\,dx^2+\left[at+\frac{b}{2}t^2+\frac{c}{3}t^3+\frac{d}{4}t^4\right]_0^x \\
&= b+2\,cx+3\,dx^2+ax+\frac{b}{2}x^2+\frac{c}{3}x^3+\frac{d}{4}x^4-0 \\
&= b+(2\,c+a)x+\left(3\,d+\frac{b}{2}\right)x^2+\frac{c}{3}x^3+\frac{d}{4}x^4
\end{align*}
This confirms that the map $T$ is, in fact, well-defined.
Note that this formula implies that the matrix of $T$ relative to the standard basis is
$$
[T]=
\begin{bmatrix}
0 & 1   & 0   & 0\\
1 & 0   & 2   & 0\\
0 & 1/2 & 0   & 3\\
0 & 0   & 1/3 & 0\\
0 & 0   & 0   & 1/4
\end{bmatrix}
$$
The reduced row-echelon form of this matrix is
$$
\DeclareMathOperator{rref}{rref}\rref[T]
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0& 1 & 0 & 0 \\
0&0&1&0\\
0&0&0&1\\
0&0&0&0
\end{bmatrix}
$$
This means that $T$ is rank four. That is, the dimension of the image of $T$ is four and a basis for the image corresponds to the span of the columns of [T]
$$
\DeclareMathOperator{Image}{Image}\Image T
=
\DeclareMathOperator{Span}{Span}\Span\left\{
x,
1+\frac{1}{2}\,x^2,
2\,x+\frac{1}{3}\,x^3,
3\,x^2+\frac{1}{4}\,x^4
\right\}
$$
