Vector Space of polynomials and the dual space with integrals Let $V = <1,x,x^2>$ be the vector space of polynomial functions $p:\Bbb R\rightarrow \Bbb R$, I am trying to show that 
$f_1(p)= \int_0^{1} p(x)dx$, $f_2(p)= \int_0^{2} p(x)dx$, and 
$f_3(p) = \int_0^{3} p(x)dx$
 is a basis of $V^*$, and to find its dual basis in $V$. 
Thoughts:
 It seems if I show that they are linearly independent, then automatically they are generating since $V^*$ has dimension $3$.
First Part (which I am more interested in):
Taking the anti derivative of a general polynomial of degree $2$ say
$a_1+a_2x+a_3x^2$, then considering a linear combination set to 0, we have by antiderivatives:
$k_1 (a_1 + a_2/2+a_3/3) + $
$ k_2 (2a_1 + 2a_2+8a_3/3) +$
$ k_3 (3a_1 + 9/2a_2+9a_3)  = 0$ for all values off $a_1, a_2, a_3$ which now amounts to showing that $k_1 = k_2 = k_3 = 0$ ?
Second Part: I have got the general idea for the second part "and to find its dual basis in V". I consider a general form $a_1+a_2x+a_3x^2$ and integrate with each function solving for coefficients where $f_1(p) = 1, f_2(p) = 0, f_3(p) = 0$ for the first dual basis of $V^*$ and switching where the $1$ value goes for the other two vectors. 
 A: Try and find the matrices of the three linear maps:
$$
f_1(1)=\int_0^1 1\,dx=1,
\quad
f_1(x)=\int_0^1 x\,dx=1/2,
\quad
f_1(x^2)=\int_0^1 x^2\,dx=1/3
$$
Thus the matrix of $f_1$ is $[1\;1/2\;1/3]$.
Similarly, the matrices of $f_2$ and $f_3$ are, respectively,
$$
[2\;1\;8/3]
\qquad
[3\;9/2\;9]
$$
and these three row vectors are linearly independent.
A polynomial such that $f_1(p)=1$, $f_2(p)=0$ and $f_3(p)=0$ must have coordinates $[a\;b\;c]^T$ such that
$$
\begin{bmatrix}
1&1/2&1/3 \\
2&1&8/3\\
3&9/2&9
\end{bmatrix}
\begin{bmatrix}a\\b\\c\end{bmatrix}=
\begin{bmatrix}1\\0\\0\end{bmatrix}
$$
and then $p=a+bx+cx^2$.
You see you just have to find the inverse of
$$
\begin{bmatrix}
1&1/2&1/3 \\
2&1&8/3\\
3&9/2&9
\end{bmatrix}
$$
A: To show it is a basis, it suffices you show that such functionals are linearly independent. Suppose then that $\sum \lambda^i f_i= 0$. If you evaluate this at the basis elements $1,2x,3x^2$ you obtain the equations
$$\lambda^1 +2\lambda^2 + 3\lambda^3=0$$
$$\lambda^1 +4\lambda^2 + 9\lambda^3=0$$
$$\lambda^1 +8\lambda^2 + 27\lambda^3=0$$
and this is $V(1,2,3)\vec \lambda =0$ where $V(1,2,3)$ is a Vandermonde matrix. Since $V(1,2,3)$ is invertible, $\vec\lambda=0$, so the set $\{f_1,f_2,f_3\}$ is linearly independent, and hence a basis because $\dim V^\ast=3$. 
To obtain a dual basis, you have to solve three systems of equations. To illustrate, if $p_i=a_0+2a_1x+3a_2x^2$ is dual to $f_i$ then
$$a_0+a_1+a_2=\delta_{1i}$$
$$a_0+4a_1+9a_2=\delta_{2i}$$
$$a_0+8a_1+27a_2=\delta_{3i}$$
so you have to invert $V(1,2,3)$. 
