Problem on dual basis 
Let V be the real vector space of all polynomials, in a single
   variable and with real coefficients, of degree at most $3$. 
Let $V^*$  be its dual space.
Let $t_1 = 1, t_2 = 2, t_3 = 3, t_4 = 4.$ 
Which of the following sets of functionals $\{f_i |1 \leq i \leq 4\}$
   form a basis for $V^*$?
a). For $1 \leq i \leq 4,$ and for all $p \in V , f_i(p) = p(t_i)$.
b). For all $p \in V , f_i(p) = p(t_i)$ for $i = 1, 2$, $f_3(p) =
 p'(t_1)$ and $f_4(p) = p'(t_2)$.  
c). For all $p \in V , f_i(p) =
 p(t_i)$ for $1 \leq i \leq 3$ and $f_4(p)=\int_{a}^{b} p'(t)dt$


For option a)  i'm doing just take  $f=af_1+bf_2+cf_3+df_4=0$. There exists a $p\in V$ with $p(1)=1,p(2)=p(3)=p(4)=0$ Since $f_i(v_j) = \delta_{ij}$. Applying $f$ to $p$ we find $a=0$, Similarly $b=c=d=0$ hence basis. but don't know how doing for b and c ????Thank you for help 

One approach is as follows: just as the linear map
$$
p \mapsto \pmatrix{p(1)\\p(2)\\p(3)\\p(4)}
$$
has a trivial kernel, show that the map
$$
p \mapsto \pmatrix{p(1)\\p(2)\\p'(1)\\p'(2)}
$$
has a trivial kernel.
For c, note that we can rewrite $f_4(p) = p(b) - p(a)$.  So, whether or not the $f_i$ are linearly independent depends on the values of $a$ and $b$.
