I have a question asking to find an orthonormal basis of $p_2$ with respect to the inner product =2 X integral from 0 to 1 p(x)q(x)dx. What do I do with the 2 in front of the integral? When I solve for v2 do I use the integral from 0 to 1 p(x)q(x)dx without multiplying by 2? Do I only multiply when solving for the norm.
In other words, using the basis $/{1,x,x^2/},$
would my $v_2$ be $x-(1/2) $ or $x-1.$
The multiplier $2$ is not effective when you look for orthogonal vectors, it effects only the norm.
For the constant polynomial $1$, we get $\|1\|^2=2\cdot\int_0^1 1\cdot 1=2$, so $\|1\|=\sqrt2$, meaning that $v_1$ should rather be the constant $\displaystyle\frac1{\sqrt2}\,$.
For $v_2$, we have $\ 2\int_0^1 1\cdot x\ =\ 1 $, so $2\int_0^1 1\cdot(x-1/2)=0$.
[But we would get the same without the $2$ in front, because $\int_0^1x$ is the half of $\int_0^11$.]
Then again, we have to norm it and finally set $$v_2:=\frac{x-1/2}{\|x-1/2\|}\,.$$