Ordinary Least Squares (OLS) + Matlab Please help me to solve the equation with Ordinary Least Squares (OLS) method.

Given two same length vectors:


$x=(x_{1} , ... , x_{n})$

$y=(y_{1} , ... , y_{n})$


1) Find coefficients a,b,c for:

$y_{i} = ax_{i}^2 + bx_{i} + c$
such that: $$\int_0^1 (ax^2+bx+c) \,dx= 0$$

2) Solve it with Matlab 

I'll be very grateful for your help!

My way:
I tried something like this:

$
\left\{
\begin{aligned}
 ax_{1}^2+bx_{1}+c &= y_{1} \\ ...\\
 ax_{n}^2+bx_{n}+c &= y_{n}
\end{aligned}\right.
$

After that:
$$ \left[
    \begin{array}{}
      x_{1}^2&x_{1}&1\\
      x_{2}^2&x_{2}&1\\...\\
      x_{n}^2&x_{n}&1\\
    \end{array}\right]\left[
\begin{array}{}
      a\\
      b\\c\\
    \end{array}
\right] = 
\left[
\begin{array}{}
      y_{1}\\
      y_{2}\\...\\y_{n}
    \end{array}
\right]$$

In Matlab:
M= [x.^2    x    ones(length(x),1)];}
U = M\Y  or U = pinv(M)*Y;

And after that:

U(1) = a;
U(2) = b;
U(3) = c;

But I don't understand why I need an Integral?
 A: Based on what you wrote it seems to me that the integral equation is imposing a constraint on your least squares solution. In order to satisfy this constraint the coefficients $a, b, c$ must satisfy the equation
$$
2a + 3b + 6c = 0,
$$
that is, solutions must lie on the plane through the origin with normal vector $(2,3,6)$. Your Matlab solution is correct in the unconstrained case, but it doesn't work in the constrained case since it finds the best solution over all of $\mathbb{R}^3$. I am not sure what Matlab toolboxes you have access to, but you may want to look at lsqlin, which solves constrained linear least-squares problems.
A: You can follow David's suggestion. As K. Miller mentioned, the integral gives you a constraint, or a relationship between the three coefficients. You can find 
$$c=-\frac{1}{3}a-\frac{1}{2}b$$
and plug this $c$ into your fitting equation. 
This gives you
$$y_i=ax_i^2+bx_i-\frac{1}{3}a-\frac{1}{2}b\\
=a(x_i^2-\frac{1}{3})+b(x_i-\frac{1}{2})$$
Now using $x_i^2-\frac{1}{3}$ and $x_i-\frac{1}{2}$ as values in your matrix, you can find the coefficients $a$ and $b$. $c$ is then determined by the first equation obtained by that integral.
