<s>[INST] How to find the orthogonal projection of the given vector on the given subspace $W$ of the inner product space $V$. $V=P(R)$ with the inner product $\langle f(x),g(x) \rangle$=$\int_0^1 f(t)g(t )dt$, $h(x)=4+3x-2x^2$ and $W=P_1(R)={\{1,x}\}$.
I don't know how to  do this the question. All I know is that it has something do with the Gram-Schmidt process.
Please provide full details ,if possible as I don't know much about about orthogonalization projection. 
 [/INST] The inner product structure of your vector space $V$ is $$\langle f | g 
\rangle = \int_0^1 f(x)g(x) dx$$
To project a vector
$$h(x) = 4 + 3x - 2x^2$$
on the subspace $W$ of the vector space $V$, you just add the projections of $h$ on each of the basis vectors of the subspace.
In this case, since 
$$W = P_1 = \left\{ 1, x \right\}$$
and the vector we wish to project is $h$, we need to find
$$
w = 1 \times\langle h |1\rangle + x \times\langle h|x\rangle 
$$ 
Where $w$ is the projection of $h$ in $W$
Let's now compute $w$
$$
w = 1 \times\langle h |1\rangle + x \times\langle h|x\rangle \\
= 1 \times \int_0^1h \cdot 1dx + x \times \int_0^1 h \cdot x dx \\
= \int_0^1 (4 + 3x - 2x^2)dx + x\int_0^1 (4 + 3x - 2x^2)x dx \\
= \int_0^1 (4 + 3x - 2x^2)dx + x\int_0^1 (4x + 3x^2 - 2x^3) dx \\
= 4x + \frac{3x^2}{2} - \frac{2x^3}{3} \bigg \vert_0^1 +
x \left( \frac{4x^2}{2} + \frac{3x^3}{3} - \frac{2x^4}{4} \bigg \vert_0^1 \right) \\
= \left(4 + \frac{3}{2} - \frac{2}{3} \right) + x \left(\frac{4}{2} - \frac{3}{3} - \frac{2}{4}\right) \\
= \frac{12 + 9 - 4}{6} + x \left(2 - 1 - \frac{1}{2} \right )\\
= \frac{17}{6} + \frac{x}{2}
$$ 
Hence, the projection of $h$ on $W$, or
$$
w = \langle h | W \rangle = \frac{17}{6} + \frac{x}{2}
$$
</s>