Given a coefficient vector for a polynomial $p(x)$ , find a matrix that calculates the coefficient vector of its anti-derivative $P(x)$ Given is a polynomial $p(x)$ of degree $n = 3$ in monomial basis:
$p(x) = \sum_{i=0}^n c_i x^i = (1,x,x^2,\ldots)\,\mathbf{c}, \, \mathbf{c} \in \mathbb{R}^{n+1}$. 
I'm trying to prove that for $p(x)$ and its coefficient vector $c$, that the coefficient vector for the polynomial's anti-derivitave $P(x)$ can be calculated using matrix multiplication. For example: $P(x): \int c = A_{T-1}c$. 
I need to find the matrix $A_{T-1}$ for integrating. When I define integration as a linear transformation and try to find $A_{T-1}$ using the standard method, I get stuck at the last basis vector $x^3$. Its anti-derivative is ${1/4} \cdot x^4$ and this cannot be represented as a linear combination of the basis vectors $\{1, x, x^2, x^3\}$. Can anyone give me a tip how I can approach this?  
Follow up question: How do the matrices $A_{T-1}$ and $A_3$ relate to each other, where $A_3$ is the matrix that calculates the derivative of the polynomials coefficient vector? 
I have calculated  $A_3$ to be: $\begin{pmatrix} 0&& 1&& 0&& 0&& \\  0&& 0&& 2&& 0&& \\  0&& 0&& 0&& 3&& \\ \end{pmatrix}$
$A_3 \cdot T_2$ gives us  $\begin{pmatrix} 0&& 1/2&& 0&& 0&& \\  0&& 0&& 2/3&& 0&& \\  0&& 0&& 0&& 3/4&& \\ \end{pmatrix}$ 
What does this mean? 
 A: Let $p(x)=c_0+c_1x+c_2x^2+c_3x^3.$ Then an anti-derivative $P$ is given by 
$$P(x)=c_0x+\frac{c_1}{2}x^2+\frac{c_2}{3}x^3+\frac{c_3}{4}x^4.$$
Now define $A:= diag (1, \frac{1}{2},\frac{1}{3},\frac{1}{4}).$ $A$ will do the job:
$$A(c_0,c_1,c_2,c_3)^T=(c_0, \frac{1}{2}c_1,\frac{1}{3}c_2,\frac{1}{4}c_3)^T.$$
A: Every antiderivative is determined up to a constant. In order to represent the linear application you describe, I will note it $T$, by a matrix, we need to set that constant. If we chose it to be $0$, then $T(\alpha)=0$ (where $\alpha$ is the constant polynomial equal to $\alpha$).
As you observed, $T(x^3) = \frac 1 4 x^4$ and then, if you work over polynomial up to degree $n$, denoted $P_n$, then the range of $T$ are polynomials up to degree $n+1$ with first coefficient being $0$, denoted as $P^0_{n+1}$.  Therefore we only need $n$ values to describe every elements in the range of $T$, $P^0_{n+1}$. Even if $T$ is not an endomorphism, you can see it as such when working only with coefficients since there is an isomorphism between $P_n$ and $P^0_{n+1}$ and they are of finite dimension.
So if we take the basis of $P_n$ to be $\{1,x^1,\dots,x^n\}$ and  $P^0_{n+1}$ to be  $\{x^1,\dots,x^n,x^{n+1}\}$, you should be able to write $A_T$, the square matrix representation of $T$ in those basis in the form that Fred described in its answer.
