<s>[INST] Finding null space and image Let $T(f(x)) = g(x) = \int_{-1}^{1}f(t)(t-x)^2dt$ be a linear transformation from $V$ to $V$ where $V$ is the vector space of continuous functions on $[-1 , 1]$ . Find $\text{Nul}(T)$ and $\text{Im}(T)$ . I really have no idea about the solution . Can $x$ be considered constant in the integral ? And then how to evaluate integral ? 
 [/INST] As other have pointed out the image of $T$ is a subspace of the $2^{nd}$ degree polynomials
To find the image let us only look at how $T$ operates on polynomials
$$ f(t) = \sum_{k=1}^n a_kt^k$$
\begin{align*}
\int_{-1}^1 f(t) (t-x)^2 dt &= \int_{-1}^1f(t)(t^2 + x^2 -2tx) \;dt \\
&= \int_{-1}^1 \sum_{k = 1}^n a_k t^{k+2} \;dt + x^2 \int_{-1}^1 \sum_{k = 1}^n a_k t^k \; dt -2x \int_{-1}^1 \sum_{k = 1}^n  a_k t^{k+1} \; dt \\
&=  \sum_{k =1}^n \frac{a_k}{k+3} (1 - (-1)^{k+3}) + x^2 \sum_{k = 1}^n \frac{a_k}{k+1} (1 - (-1)^{k+1}) -2x \sum_{k = 1}^n \frac{a_k}{k+2} (1-(-1)^{k+2}) \\
&= \sum_{k =1 \\k \text{ even}}^n \frac{2a_k}{k+3} + x^2 \sum_{k = 1 \\ k \text{ even}}^n \frac{2a_k}{k+1} +x \sum_{k = 1 \\ k \text{ odd}}^n \frac{-4a_k}{k+2} \\
&= \Big( \sum_{k = 1 \\ k \text{ even}}^n \frac{2a_k}{k+1} \Big) x^2 + \Big( \sum_{k = 1 \\ k \text{ odd}}^n \frac{-4a_k}{k+2}\Big) x + \sum_{k =1 \\k \text{ even}}^n \frac{2a_k}{k+3}
\end{align*}
Notice that this is actually a polynomial in $x$. So $T(f)(x) = ax^2 +bx +c$ is actually a linear system of equations in $a_1,a_2,...a_n$
Also notice the equations are linearly independent for $n$ greater than $4$.
If $n = 4$ and the variables $a_1 = x, a_2 = y, a_3 = z, a_4 =t$ we have the following system
Let $a,b,c \in \mathbb{R}$
\begin{align*}
\frac{2}{3} y + \frac{2}{5} t &= a \\
\frac{-4}{3} x +  \frac{-4}{5} z&= b \\
\frac{2}{5} y + \frac{2}{7} t  &= c
\end{align*}
By solving the two equations in $y$ and $t$  first we get a unique solution for $y$ and $t$. Then we can always chose $x$ and $z$ such that the last equation is satisfied.
Therefore 
 $$ Im(T) = \{f(x) \in \mathbb{R}[X] : deg(f)\leq 2\}.$$
</s>