Im struggling with the following Let F: $\displaystyle \P_2(R)$ $\displaystyle \rightarrow$ $\displaystyle \P_3(R)$ be the linear transformation determined by: F(f)(x)=6$\displaystyle \int_{0}^{2x-1} f(t) dt$

compute the dimension of the kernel of F. Determine a basis for the image of F.

Firstly is the dimension of ker(F)=1,

Then I have found F(f)(x) and have that it equals: 16ax^3+(12b-24a)x^2 +(12a-12b+12c)x + (-2a+3b-6c), from this is our basis given by [(-2a+3b-6c),(12a-12b+12c),(-24a+12b),16a] ?

Many thanks in advance, any help would be most appreciated.