well, a little thought should convince you that we can't "transform" a smaller space into a bigger one (in a linear way), we can only (at best) place a "copy" of our smaller space inside the bigger one.

the number-one tool to use in these situations is the rank-nullity theorem, so dim(P_{2}) = dim(ker(F)) + dim(im(F)).

since dim(P_{2}) = 3, even if dim(ker(F)) = 0 (the smallest possible value), the biggest dim(im(F)) could possibly be is 3. this means that {1,x,x^{2},x^{3}} couldn't possibly be a basis, as that set has 4 elements (4 is bigger than 3, see? this vector space stuff is easy).

since it's not really clear how big a space we get by taking that integral, let's look at the kernel first. at least this will tell us how many basis vectors we have to come up with for the image.

now ker(F) consists of all the polynomials of degree 2 or less (plus the 0-polynomial, which doesn't really have a degree) that F takes to the 0-polynomial in P_{3}.

so if f(x) is in ker(F):

F(f)(x) = 0 (= 0 + 0x + 0x^{2}+ 0x^{3}, NOT "the real number 0").

so we're going to have to bite the big one, and actually do some integration, here (i hope i can remember HOW). let's take a "typical polynomial in P_{2}", say:

f(x) = a + bx + cx^{2}, and just "see what happens". so

(whew! i hope i did that right. you might want to double-check it).

since this is identically 0, this gives us the four equations:

26c/3 = 0

4b + 10c = 0

2a + 4b + 2c = 0

2a + 2c/3 = 0

from which we see that c = 0, and thus a and b must also be 0. hence dim(ker(F)) = 0, so F is injective. this tells us that we can form a basis for im(F), by taking the images of any basis for P_{2}. and we happen to have one handy: {1,x,x^{2}}. so it suffices to calculate F(1)(x), F(x)(x), and F(x^{2})(x) (those look kind of funny, don't they? perhaps we should have chosen another variable to use for P_{2}. too late, we're forging ahead).

now:

so {2x+2,4x^{2}+4x,(26/3)x^{3}+10x^{2}+2x+(2/3)} is a basis for im(F).