My question is this:
Say that the set H is the set of polynomials $p \in P_2$ which are in the form:
$$ p(x) = \int_1^x q(t)dt $$
for some other polynomial $q$.
I need to show that,
a) H is a subspace of $P_2$ and
b) assuming that $T : H \to R$ is a function defined by $T(p) = p'(2)$, show that T is a linear transformation.
For a), since $p \in P_2$, any element in p can be defined as $p(x) = a_0 + a_1x + a_2x^2$, where $a_0, a_1, a_2$ are arbitrary constants. q would have to be defined as (I guess it has to be a polynomial of any degree?) $q(t) = b_0 + b_1t + b_2t^2 + … +b_nt^n$. Before I take another step, I'd like to know, would I have to show that
$$\int_1^x b_0 + b_1t + b_2t^2 + … +b_nt^n \to (b_ot + \frac{b_1t^2}{2} + \frac{b_2t^3}{3}+…+ \frac{b_nt^{n+1}}{n+1}) |^x_1 $$
contains the zero vector, and is closed under multiplication and addition? If so, how? Or have I overthought this?
For b), well, $p'(x) = a_1 + 2a_2x$ and therefore $p'(2) = a_1 + 4a_2$. I'm not certain where to go from here; it does make sense, since $a_1 + 4a_2$ does always map to a real number, but that is not exactly a "proof".
Best Answer
One mistake you make is assuming "any degree". Here we are at degree at most $2$.
Because any degree-zero polynomial satisfies $f(x) = f(1)$, $$0 = \int_1^x 0 \ dt$$
Sums and scalar multiples follow from linearity of the definite integral.
For showing $T(p) = p'(2)$ is linear:
$T(ap(t) + bq(t)) = ap'(t) + bq'(t) \Big|_{t=2} = ap'(2)+bq'(2) = aT(p)+bT(q)$
The first equality is by linearity of differentiation. Notice the fact that they were polynomials wasn't even necessary, just that they were functions differentiable at $2$. If you want to be safe and define arbitrary polynomials,
$$p(t) = a_0+a_1t+a_2t^2, \ \ q(t) = b_0 + b_1t + b_2t^2$$
then go ahead and differentiate them, plug in $2$, and show it is the same for the polynomial $(p+q)(t) = (a_0+b_0) + (a_1+b_1)t + (a_2+b_2)t^2$.