Let's us use the standard basis for $P_1$, the vector space of polynomials of degree at most $1$. We will define the mapping $T(p)=\int_0^x p(t)\ dt$ notice that we need to take a definite integral to avoid ambiguity in the constant of integration.
Let us first prove that this mapping is linear. If we let $p$ and $q$ be polynomials in $P_1$ with scalar $c$, then we have $T(cp + q) = \int_0^x cp(t) + q(t)\ dt = c\int_0^x p(t)\ dt + \int_0^x q(t)\ dt = cT(p) + T(q)$ so the mapping is indeed linear.
If we feed the standard basis vectors into the mapping, we end up with $T(1) = \int_0^x 1\ dt = x$ $T(t) = \int_0^x t\ dt = \frac{x^2}{2}$ We can write the matrix of the mapping $T$ with respect to the standard basis vectors of $P_1$ and $P_2$ as $[T] = \begin{pmatrix}0 & 0 \\ 1 & 0 \\ 0 & \frac{1}{2}\end{pmatrix}$