Linear transformation with an integral. Define $T: \Bbb{R}_2[x] \to M_2(\Bbb{R})$ by
$$
T(p(x)) := \begin{pmatrix}
p(\beta+1) & p'(\alpha)\\
\int_{-1}^1 p(x) \ dx & p(-\beta)
\end{pmatrix} 
$$
where $\alpha=1$ and $\beta=0$.
I'd like to prove that $T$ is a linear transformation and find $\mathrm{im}(T)$ and $\ker(T)$.
I'm going to be honest with you all, I don't have any idea on how to solve this question because I'm in calculus I, so I don't know anything about integration.
If someone knows how to solve this, please help me if you want to.
 A: Here's a sketch of the solution. I'll leave some details for you to verify.
From what was discussed in the comments, it should be clear that $T$ is a linear map.
Any element in $\Bbb{R}_2[x]$ looks like $p(x):=ax^2+bx+c$ for some $a,b,c \in \Bbb{R}$. Thus as a vector space $\Bbb{R}_2[x]$ is $\Bbb{R}^3$ in disguise.
Using that $\alpha=1$ and $\beta=0$, you should quickly get that $p(1)=a+b+c$, $p(0)=c$, $p'(1)=2a$ and
$$ 
\int_{-1}^1p(x)\ dx=\frac{2}{3}a+2c.
$$
Therefore, the map $T$ (regarded as a map from $\Bbb{R}^3$ to $M_2(\Bbb{R})$) looks like
$$
T(a,b,c) := \begin{pmatrix}
a+b+c & 2a\\
\frac{2}{3}a+2c & c
\end{pmatrix} 
$$
From here is easy to see that $\mathrm{ker}(T)=\{(0,0,0)\}$, which in turn means that the only element in $\ker(T)$ is the $0$ polynomial (whence $T$ is injective). To find $\mathrm{im}(T)$ observe that
$$
\begin{pmatrix}
a+b+c & 2a\\
\frac{2}{3}a+2c & c
\end{pmatrix} = a\begin{pmatrix}
1 & 2\\
\frac{2}{3} & 0
\end{pmatrix} + b \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}+ c \begin{pmatrix}
1 & 0\\
2 & 1
\end{pmatrix}.
$$
From here, you should be able to deduce that $\mathrm{im}(T)$ is the span of the three (linearly independent) matrices shown on the right hand side of the last equation above.
