a problem in linear transformation The problem says to show that the mapping $T:P(R)->P(R)$ defined by
$$T(f(x))=\int_0^x f(t) \ dt$$
is a linear transformation which is injective  but not surjective(Friedberg BOOK,PAGE-75,problem no -15)
Now my question is that why $T$ is acting on $f(x)$(i.e the image) instead of acting on the member of $P(R)$ i.e on the function $f$ and what is the role of $x$ in the definition of $T$(i.e,my point is to understand the definition of $T$ explicitly).
secondly,assuming the definition(without understanding it properly) i have shown that $T$ is linear but i got stuck in showing that $T$ is injective(tried to show that $Ker\ T=\{0\}$ but failed) and most importantly not surjective(in proving not surjective i am not seeing any clue)
any kind of help to this problem is welcome.
 A: While $P(\mathbb{R})$ is a codomain, it is also a domain of $T$, so maybe you are confusing yourself a bit. $f(x)$ may be an image, and but it is also a member of $P(R)$.
The transformation $T$ is mapping a polynomial function $P(\mathbb{R})$ to another polynomial function in $P(\mathbb{R})$. For instance:
$x+1 \in P(\mathbb{R})$ and $\displaystyle T(x+1)=\int_{0}^{x} t+1 \, dt = \frac{x^2}{2} +x \in P(\mathbb{R}) $
This is linear since integration is clearly a linear mapping.
For injectivity, suppose that $T(f)=0$. Since $f$ is a polynomial with real coefficients, you can definitely express it in explicit form
$$
f(x)=\sum_{k=0}^{n} a_k x^k
$$
Then what is $T(f)$? What does the condition $T(f)=0$ tell you about coefficients of $f$?
A: $P(\mathbb R)$ denotes the set of all polynomial with real coefficients.
i.e., $P(\mathbb R)=\{f(x)\  |\ f(x)= a_nx^n+a_{n-1}x^{n-1} +\dots +a_1x^1+a_o, a_i \in \mathbb R\}.$
Here,  $T(f(x))=\int_0^x \ f(t) \ dt = \int_0^x \sum_{i=0}^n a_ix^i= \sum_{i=0}^n \frac{a_i}{i+1}x^{i+1}$
$ \because T(f(x)+g(x))= \int_0^x(f(t)+g(t))\ dt=\int_0^x f(t)\ dt + \int_0^x g(t)\ dt =T(f(x))+T(g(x))$,  we have $T$ is linear.
Suppose $Ker\ T \neq \{0\}.$ Then there is a non zero polynomial $f(x)$ in $P(\mathbb R)$ such that $T(f(x)) = 0.$ 
$\implies \int_0^xf(x)=0 \implies f(x)=0,$ a contradiction. Hence $Ker\ T =\{0\}.$
Consider a constant polynomial $f(x) = k $ for some $k \in \mathbb R.$ There is no $f(x)\in P(\mathbb R)$ such that  $\int_0^x f(x) =k.$ Therefore constant polynomials does not have pre images, hence $T$ is not onto.
