Finding the min of an integral So I have to find the following $$\min_{a,b,c\in\mathbb{R}}\int_{-1}^{1} |x^3-a-bx-cx^2|^2dx$$
I have a hint at a solution which says to consider $X=\{\mbox{polynomials of degree} \leq 2\}$.
So then we have $$\min_{a,b,c\in\mathbb{R}}\int_{-1}^{1} |x^3-a-bx-cx^2|^2dx=\inf_{g\in X} ||x^3-g||$$ for some $g\in X$
Where the norm $||.||$ is defined using the inner product $<f,g>=\int_{-1}^{1} f(x)\bar{g(x)}dx$
So then I think I'm supposed to use an orthogonal projection somehow I think (and maybe find some orthonormal basis?) but I'm a bit lost as to how to do any of this.
Thanks for any help.
 For completeness I am putting my solution (from the answers below).

The problem reduces to finding a basis for $X$ and then orthonarmalizing it. We then use the fact that the orthohgonal projection onto the space $X$ will give the minimal distance to it so we need to calculate:
$$||x^3-P(x^3)||$$ where $P(x^3)=\sum_{i=1}^{3} <x^3,e_i>e_i$ for an orthonormal basis $\{e_i\}$.
Noting that the set $\{1,x,x^2\}$ spans the space $X$ we then apply the gram-schmidt process to this set to give a set of orthogonal vectors: $$\{1,x,(x^2-\frac{1}{3})\}$$. 
This basis now needs to be normalized but if we notice that in :
$$\sum_{i=1}^{3} <x^3,e_i>e_i$$ 
The first and third terms will cancel as the integrand will be odd, so we only need to normailze the middle vector which gives $\{\sqrt{\frac{3}{2}\}}$.
And so $$P(x^3)=<x^3,\sqrt{\frac{3}{2}}x>\sqrt{\frac{3}{2}}x=\frac{3}{5}x$$
So we now have $||x^3-\frac{3x}{5}||=\sqrt{\int_{-1}^{1} (x^3-\frac{3x}{5})^2dx}=\sqrt{\frac{8}{175}}$
Which is the desired distance
 A: $$
\begin{eqnarray}
\min f\left(a,b,c\right) &=& \min\int_{-1}^1 dx \left(x^3 - a - bx - cx^2\right)^2 \\
&=& 2 \min \left[\left(a^2 + \frac{2}{3}ac + \frac{1}{5}c^2\right) + \left(\frac{1}{3}b^2 - \frac{2}{5}b\right) + \frac{1}{7}\right] \\
&=& 2 \min \left[g\left(a,c\right) + h\left(b\right) + \frac{1}{7}\right] \\
&=& 2 \left[g\left(0,0\right) + h\left(- \frac{3}{25}\right) + \frac{1}{7}\right] \\
&=& \frac{8}{175}.
\end{eqnarray}
$$
A: In fact you need to find square of the distance from the vector $x^3$ to the subspace spanned by $1,x,x^2$ in innner product space of polynomials on $[-1,1]$.
Here is a big picture approach. Let $H$ be a inner product space, $X$ its finite dimensional subspace with orthogonal basis $\{e_1,\ldots,e_n\}$. Let $x\in H$, then the orthogonal projection of $x$ onto $X$ is given by
$$
\mathrm{pr}_X(x)=\sum\limits_{k=1}^n \langle x,e_k\rangle e_k\tag{1}
$$
The dsired distance is
$$
\mathrm{dist}(x,E)=\Vert x-\mathrm{pr}_E(x)\Vert\tag{2}
$$
In your case take $H$ to be the space of all polynomials on $[-1,1]$, and $X$ to be the space of all polynomials of degree less than $2$. In order to apply formula $(2)$ you need orthogoanl basis of $X$. To get it apply Hilbert-Schmidt orthogonalization process to vectors $1,x,x^2$. 
So it is remains to perform all calculations.
