Problem goes as follows:
Let $\mathbb{R}_{1}[x]$ be the linear space of polynomials of degree $\leq 1$. Define the covectors $f_{1}, f_{2} \in (\mathbb{R}_{1}[x])^{*}$ (dual space), as:
$$ f_{1}(p(x)) = \int_{0}^{1} p(x) \,dx $$
$$ f_{2}(p(x)) = \int_{0}^{2} p(x) \,dx $$.
Determine the basis of $\mathbb{R}_{1}[x]$ whose dual basis is $\{ f_{1}, f_{2} \}$.
My solution:
Let $\{v_{1}, v_{2}\}$ be the basis dual to $\{ f_{1}, f_{2} \}$.
Given the standard basis of $\mathbb{R}_{1}[x]$, $\{ e_{1} = 1, e_{2} = x \}$, I can form a matrix $f_{i}(e_{j})$, evaluating, I determine it to be:
$$ \begin{pmatrix} 1 & \dfrac{1}{2} \\ 1 & 2 \end{pmatrix} $$
The inverse of this matrix is
$$ \begin{pmatrix} \dfrac{4}{3} & -\dfrac{1}{3} \\ -\dfrac{2}{3} & \dfrac{2}{3} \end{pmatrix} $$
So the columns of this matrix are the coefficients of the polynomials I looking for. Hence,
$v_{1} = \dfrac{4}{3} - \dfrac{2}{3}x, v_{2} = -\dfrac{1}{3} + \dfrac{2}{3}x$
Is this correct?
You have that $f_1(a+bx)=a+\tfrac1{2}b$ and $f_2(a+bx)=2(a+b)$, therefore $f_1(a+bx)=1$ and $f_1(a+bx)=0$ if and only if $a=2$ and $b=-2$. By the other hand $f_1(a+bx)=0$ and $f_2(a+bx)=1$ if and only if $b=1$ and $a=-\tfrac1{2}$, so the dual basis is $\{2-2x,-\tfrac1{2}+x\}$.