What is the Coefficient Matrix of $T(p(t))=\int_0^t\int_0^yp(x)dxdy$ that maps $P_3\rightarrow P_5$? The usual basis for $P_n$, of course, is given by $\left\{1,t,t^2,\cdots,t^n\right\}$. Why is the integrand a function of $x$? Does this matter for the purposes of constructing a change of basis matrix? How do I represent integration with respect to $y$ in either basis in terms of the components of a matrix?
 A: Interchanging the order of integration allows us to write $T$ as a single integral:
$$ T(p(x))(t) = \int_0^t \int_0^y p(x) \, dx \, dy = \int_0^t p(x) \int_x^t \, dy \, dx = \int_0^t (t-x)p(x) \, dx $$
(the region of integration is $0<x<y<t$, which tells you what the new limits should be)
Now, given $p(x)=x^n$, we find
$$ T(x^n)(t) = \int_0^t (t-x)x^n \, dx = \left[ t\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^t = t^{n+2} \left( \frac{1}{n+1}-\frac{1}{n+2} \right) = \frac{t^{n+2}}{(n+1)(n+2)}, $$
so you get 
$$
\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 \frac{1}{2} & 0 & 0 & 0 \\
 0 & \frac{1}{6} & 0 & 0 \\
 0 & 0 & \frac{1}{12} & 0 \\
 0 & 0 & 0 & \frac{1}{20} \\
\end{array}
\right),
$$
I suppose.
A: The $x$ and $y$ are dummy variables, they don't really matter. For example, if $p(t) = t - 1$, we have $p(x) = x - 1$, so that
\begin{align*}\int_0^t\int_0^y (x - 1)\ dx\ dy &= \int_0^t\left[\frac{x^2}{2} - x\right]_0^y\ dy\\
&=\int_0^t\left(\frac{y^2}{2}-y\right)\ dy \\
&=\left[\frac{y^3}{6} - \frac{y^2}{2}\right]_0^t\\
&=\frac{1}{6}t^3 - \frac{1}{2}t^2.
\end{align*}
Perhaps somebody smarter than me knows how to compute the final transformation matrix from smaller matrices (ones involving integration), but I'd just bite the bullet and compute the integral for each basis polynomial in $P_3$.
You may be able to find the transformation matrix for a single integration, and apply that twice to your basis vectors, but I'm not positive it will work out.
