Let  {jatex options:inline}V{/jatex}  be the vector space of polynomials of degree 2 and let  {jatex options:inline}\phi_1 , \phi+2 : V \rightarrow \mathbb{R}{/jatex}  be linear functionals on  {jatex options:inline}V|{/jatex}  defined by
{jatex options:inline}\phi_1 (\mathbf{v})= \int^1_0 (a+bx)dx{/jatex}
{jatex options:inline}\phi_2 (\mathbf{v})= \int^2_0 (a+bx)dx{/jatex}
We want to find a basis  {jatex options:inline}\{ \mathbf{v}_1, \mathbf{v}_2 \}{/jatex}  of  {jatex options:inline}V{/jatex}  dual to  {jatex options:inline}\{ \phi_1 , \phi_2 \}{/jatex}.
The vector space  {jatex options:inline}V{/jatex}  has dimension 2 since  {jatex options:inline}\{1,x\}{/jatex}  forms a basis. The set of all linear functions from  {jatex options:inline}V{/jatex}  to  {jatex options:inline}\mathbb{R}{/jatex}  forms a vector space of dimension 2. The dual vector space  {jatex options:inline}V^*{/jatex}  spanned by  {jatex options:inline}\{ \phi_1 , \phi_2 \}{/jatex}  has dimension 2 since  {jatex options:inline} \phi_1 , \phi_2 {/jatex}  are linearly independent so form a basis for  {jatex options:inline}V^*{/jatex}.
We can use this basis to find the dual basis for  {jatex options:inline}V{/jatex}  using the following theorem.
Let  {jatex options:inline}\{ f_1, \; f_2 \}{/jatex}  be a basis for  {jatex options:inline}V^*{/jatex}  and let  {jatex options:inline}\{ \mathbf{e}_1, \; \mathbf{e}_2 \}{/jatex}  be a basis for  {jatex options:inline}V{/jatex}  then
{jatex options:inline}f_i (\mathbf{e}_j)= \left\{ \begin{array}{cc} 1 & i=j \\ 0 & i \neq j \end{array} \right. {/jatex}
Let  {jatex options:inline}\mathbf{e}_1=a_1+b_1x, \; \mathbf{e}_2=a_2+b_2x {/jatex}. Then
{jatex options:inline}\phi_1 (\mathbf{e}_1)= \int^1_0 (a_1+b_1x)dx=a_1+\frac{b_1}{2}=1{/jatex}
{jatex options:inline}\phi_2 (\mathbf{e}_1)= \int^2_0 (a_1+b_1x)dx=2a_1+2b_1=0{/jatex}
{jatex options:inline}\phi_1 (\mathbf{e}_2)= \int^1_0 (a_2+b_2x)dx=a_2+\frac{b_2}{2}=0{/jatex}
{jatex options:inline}\phi_2 (\mathbf{e}_2)= \int^2_0 (a_2+b_2x)dx=2a_2+2b_2=1{/jatex}
Solving these equations for  {jatex options:inline}a_1, \; b_1. \; a_2, \; b_2{/jatex}  gives  {jatex options:inline}a_1=2, \; b_1=-2, \; a_2=-1, \; b_2=2{/jatex}. Hence the basis dual to  {jatex options:inline}\{ \phi_1, \; \phi_2 \}{/jatex}  is  {jatex options:inline}\{ 2-2x, \; -1+2x{/jatex}.