Let's start with an image ($k=3$, WENO5 scheme)
The polynomials $p_r(x), r = 0, \dots, k-1$ defined each on its stencil $S_r$ are obtained by solving the following systems of equations
$$
\frac{1}{\Delta x}\int_{x_{j-1/2}}^{x_{j+1/2}} p_r(x) dx = v_{j}, \qquad j = i-r, \dots, i-r+k-1
$$
Here $v_j$ denotes the cell-averaged value of $v(x)$ in $[x_{i-1/2}, x_{i+1/2}]$. We're trying to reconstruct $v(x)$ on $S_r$ from its cell averages as $p_r(x)$.
Each system has $k$ equations and $k$ unknown coefficients in $p_r(x)$.
To simplify equations a bit I will use dimensionless form of the polynomials' argument:
$$
p_r(x) = \tilde p_r\left(\frac{x - x_i}{\Delta x}\right)\\
\tilde p_r(\xi) = p_r(x_i + \xi \Delta x)
$$
The equations become clearer:
$$
\int_{j-1/2}^{j+1/2} \tilde p_r(\xi) d\xi = v_{i+j}, \qquad j = -r, \dots, k-r-1
$$
So do the smoothness indicators $\beta_r$:
$$
\beta_r = \int_{-1/2}^{1/2} [\tilde p_r'(\xi)]^2 + [\tilde p_r''(\xi)]^2 + \dots +
[\tilde p_r^{(k-1)}(\xi)]^2 d\xi
$$
and the values at the interfaces $v_{i+1/2}^{(r)} = \tilde p_r(1/2)$.
Stencils $S_0, \dots, S_{k-1}$ differ only by shift. This implies that $p_r$ are not completely independent. Indeed, let's compare equations for $\tilde p_0$ and $\tilde p_r$:
$$
\int_{j-1/2}^{j+1/2} \tilde p_0(\xi) d\xi = v_{i+j}, \qquad j = 0, \dots, k-1\\
\int_{j'-1/2}^{j'+1/2} \tilde p_r(\xi) d\xi = v_{i+j'}, \qquad j' = -r, \dots, k-r-1
$$
Plugging $j' = j-r$ we get
$$
\int_{j-r-1/2}^{j-r+1/2} \tilde p_r(\xi) d\xi = v_{i+j-r}, \qquad j = 0, \dots, k-1
$$
Now change $\eta = \xi + r$
$$
\int_{j-1/2}^{j+1/2} \tilde p_r(\eta - r) d\eta = v_{i-r+j}, \qquad j = 0, \dots, k-1
$$
We obtained pretty obvious (from the image above) property
$$
\tilde p_r(\xi) = \tilde p_0(\xi + r)\Big|_{v_i \mapsto v_{i-r}}
$$
All $\beta_r$ can be expressed in $\tilde p_0$ only:
$$
\beta_r = \left.\int_{r-1/2}^{r+1/2} [\tilde p_0'(\xi)]^2 + [\tilde p_0''(\xi)]^2 + \dots + [\tilde p_0^{(k-1)}(\xi)]^2 d\xi\right|_{v_i \mapsto v_{i-r}}.
$$
So are the values at the interfaces $v_{i+1/2}^{(r)} = \left.\tilde p_0(r + 1/2)\right|_{v_i \mapsto v_{i-r}}$.
Now we can finally focus on finding the exact form of $\tilde p_0(\xi)$.
Approach 1. Brute force. Simply let $\tilde p_0(\xi) = c_0 + c_1 \xi + \dots + c_{k-1} \xi^{k-1}$. Equations
$$
\int_{j-1/2}^{j+1/2} \tilde p_0(\xi) d\xi = v_{i+j}, \qquad j = 0, \dots, k-1 \tag{*}
$$
are basically a system of $k$ linear equations for $k$ unknowns $c_m$. This gives the following form for $\tilde p_0(\xi)$:
$$
\tilde p_0(\xi) =
\begin{pmatrix}
1 & \xi & \cdots & \xi^{k-1}
\end{pmatrix}
A^{-1}
\begin{pmatrix}
v_i\\
v_{i+1}\\
\vdots\\
v_{i+k-1}
\end{pmatrix}
$$
where entries of $A$ are given by
$$
a_{jm} = \int_{j-1/2}^{j+1/2} \xi^m d\xi = \frac{(j+1/2)^{m+1} - (j-1/2)^{m+1}}{m+1}, \qquad j,m = 0, \dots, k-1.
$$
Example for $k = 3$:
$$
A = \begin{pmatrix}
1 & 0 & 1/12\\
1 & 1 & 13/12\\
1 & 2 & 49/12\\
\end{pmatrix}, \quad
A^{-1} = \frac{1}{24}\begin{pmatrix}
23 & 2 & -1\\
-36 & 48 & -12\\
12 & -24 & 12
\end{pmatrix}
$$
$$
\tilde p_0(\xi) = \frac{23 v_i + 2v_{i+1} - v_{i+2}}{24} +
\frac{-3 v_i + 4v_{i+1} - v_{i+2}}{2} \xi +
\frac{v_i - 2v_{i+1} + v_{i+2}}{2} \xi^2.
$$
Approach 2. Reducing to the interpolation problem. Consider the antiderivative $P(\xi) = \int \tilde p_0(\xi) d\xi$. Rewriting the equations (*) using $P(\xi)$ gives
$$
P(j+1/2) - P(j-1/2) = v_{i+j}, \qquad j = 0, \dots, k-1.
$$
Function $P(\xi)$ is a polynomial of degree $k$ and has $k$ constraining equation. One more equation may be imposed. Let's use $P(-1/2) = 0$. The system becomes
$$
P(-1/2) = 0\\
P(1/2) - P(-1/2) = v_{i}\\
\vdots\\
P(k-1/2) - P(k-3/2) = v_{i+k-1}\\
$$
Summing the $j+1$ first equation gives
$$
P(j-1/2) = \sum_{m=0}^{j-1} v_{i+m}, \quad j = 0, \dots, k.
$$
This is clearly an interpolation problem now: find a polynomial $P(\xi)$ of degree $k$ by its known values $P(j-1/2) = V_j = \sum_{m=0}^{j-1} v_{i+m}$. The $P(\xi)$ might be expressed using Newton's interpolation formula
$$
P(\xi) = 0 + v_i (\xi + 1/2) + \frac{v_{i+1} - v_{i}}{2} (\xi + 1/2) (\xi - 1/2) + \dots {} \\ {} \dots + [V_0, \dots, V_k] (\xi + 1/2) \cdots (\xi - k + 3/2).
$$
Example for $k = 3$. The diveded difference table:
$$
\begin{array}{c|cccccc}
-1/2 & 0\\
&& v_i\\
1/2 & v_i&&\frac{v_{i+1} - v_i}{2}\\
&& v_{i+1} && \frac{v_{i+2} - 2v_{i+1} + v_i}{6}\\
3/2 & v_i + v_{i+1}&&\frac{v_{i+2} - v_{i+1}}{2}\\
&& v_{i+2}\\
5/2 & v_i + v_{i+1} + v_{i+2}\\
\end{array}
$$
$$
P(\xi) = v_i (\xi + 1/2) + \frac{v_{i+1} - v_i}{2} (\xi + 1/2) (\xi - 1/2) + {} \\ {} + \frac{v_{i+2} - 2v_{i+1} + v_i}{6} (\xi + 1/2) (\xi - 1/2) (\xi - 3/2)
$$
$$
\tilde p_0(\xi) = P'(\xi) = v_i + (v_{i+1} - v_i)\xi + \frac{v_{i+2} - 2v_{i+1} + v_i}{24} (12\xi^2 - 12\xi - 1).
$$
P.S. The computations become much more complex with large $k$ values so I suggest using some computer algebra system instead of doing computations manually.
