dimension of the image of a linear transformation of polynomials 
Let $V$ be the vector space of the polynomials of degree les or equal than $3$. Let $\Omega$ be the linear transformation such as
\begin{matrix}
\Omega: V &\rightarrow& \mathbb{R}^4\\ p &\mapsto& \left(p(a),p(b),p(c), \int_a^b p(t)\,dt\right)
\end{matrix}
  with $c\in(a,b)$.
Determine $dim(Im(\Omega))$ as a function of $c$ and show that exists $\alpha,\beta,\gamma\in\mathbb{R}$ such as
  $$\int_a^bp(t)dt = \alpha p(a)+\beta p(b) + \gamma p\left(\frac{a+b}2\right)$$

I prove that $dim(Im(\Omega)) = \begin{cases}3 &\text{if $c=(a+b)/2$}\\4&\text{if $c\neq(a+b)/2$}\end{cases}$
calculating the determinant of the matrix associated to the transformation. But I would like to know a shorter way to do it (because the arithmetic is a little bit tedious)
 A: I think you can make the problem easier by first transforming $t$. Sending $t$ to $\frac{(b-a)t+(b+a)}{2}$ gives an isomorphism of $V$ so we can assume $a=-1$ and $b=1$. Now it is easier to check that the image of $1,t,t^2,t^3$ is independent if and only if $c\neq\frac{a+b}{2}$.  
That was the important reduction. The rest is just some calculation but its "in your head" calculation rather than computing determinants. 
For example. It's clear the images are dependent when $c=0$ since then the image of $t$ and $t^3$ are both $(-1,1,0,0)$.
otherwise, subtracting the image of $t$ from the image of $t^3$ we see $(0,0,1,0)$ is in the image. Subtracting the image of $t^2$ from the image of $1$ and using that $(0,0,1,0)$ is in the image we see that $(0,0,0,1)$ is in the image. Then add the image of $t$ to that of $1$ and use that $(0,0,0,1)$ and $(0,0,1,0)$ are in the image to see $(0,1,0,0)$ is in the image. Finally subtract the image of $t$ from that of $1$ to see that $(1,0,0,0)$ is in the image. 
