Let $T:P_2 \to R$ be given by $T(p(t))=\int_{-2}^{3}p(t)dt$ please help, this linear transformation has me completely confused!
Let $T:P_2 \to R$ be given by $T(p(t))=\int_{-2}^{3}p(t)dt$
a.) prove that $T$ is a linear transformation. - For this i know that i need to show that $T$ preserves vector addition and scalar multiplication, but i do not know how to show this abstractly, i mean i have know idea how to set up my vectors.
b.) find the kernel of $T$. - i know that i need to solve for $T(p(t)) = 0$, but again the integral is confusing me
c.) find the nullity of $T$ and the rank of $T$ - only way i know to find these is with the rre form of the standard matrix representation (SMR), which im not sure how to get
d.) is $T$ an isomorphism? - I know that $T$ must be one-to-one and onto if it is an isomorphism, which can be shown with rank and nullity or if the $T(v) = 0 \implies v = 0 $ right?
I think the integral is really confusing me and the fact that $T$ is a mapping from $P_2 \to R$ its just not making sense to me right now, any help is greatly appreciated! thanks in advance. 
 A: a) The integral is linear, so $T$ is linear too :
$$
\int_{-2}^3(ap_1(t)+bp_2(t))\,dt=a\int_{-2}^3p_1(t)dt+b\int_{-2}^3p_2(t)\,dt
$$
b) Your polynomials are in $P_2$, so in general $p(t)=at^2+bt+c$ for some coefficients $a,b,c\in\mathbb{R}$. Then
$$
\int_{-2}^3(at^2+bt+c)\,dt=0
$$
is equivalent to (verify this)
$$
14a+3b+6c=0\tag{1}
$$
So, the kernel is $\{at^2+bt+c\in P_2:14a+3b+6c=0\}$
c) The nullity of $T$ is the dimension of $\ker T$. Note that $(1)$ gives you two degrees of freedom, e.g. $a$ and $b$ can be chosen to be whatever real numbers you want, and $c$ is then
$$
c=-\frac{14}{6}a-\frac{1}{2}b
$$
So, you can write $\ker T$ as
$$
\left\{p(t)\in P_2:p(t)=a\left(t^2-\frac{14}{6}\right)+b\left(t-\frac{1}{2}\right)\text{ for some }a,b\in\mathbb{R}\right\}
$$
or
$$
\ker T=\text{Span}\left\{t^2-\frac{14}{6},t-\frac{1}{2}\right\}
$$
where $t^2-\frac{14}{6}$ and $t-\frac{1}{2}$ are obviously linearly independent. Hence, $\dim\ker T=2$, that is to say, the nullity of $T$ is $2$.
Since $\dim P_2=3$, the Rank-Nullity Theorem tells you that the rank of $T$, that is, the dimension of the image of $T$, is $3-2=1$.
d) Since $\ker T\neq\{0\}$, $T$ is not injective. Hence $T$ is not an isomorphism.
A: Let $p(t)$ and $q(t) \in P_2$ and k $\in R$.
$$T\big (p(t) + kq(t)\big ) = \int_{-2}^{3} \big (p(t) + kq(t)\big )dt$$
$$=\int_{-2}^{3}p(t)dt + k\int_{-2}^{3}q(t)dt $$
$$= T\big(p(t)\big) + kT\big(q(t)\big)$$
$\implies T$ is a linear transformation.
$$N\big(T\big) = \big\{p(t) \mid T\big(p(t)\big) = 0\big\} $$
$$\iff \int_{-2}^{3}p(t)dt = 0$$
$$\iff \int_{-2}^{3}(at^2 + bt + c)dt = 0 $$
$$\iff \frac{a}{3}(3^3 - (-2)^3) + \frac{b}{2}(3^2 - (-2)^2) + c(3 -(-2))= 0$$
$$\iff  \frac{35a}{3} +  \frac{5b}{2} + 5c = 0$$
$$\iff c =  -\frac{7}{3}a -  \frac{1}{2}b$$
Thus it is all polynomials of the form $$p(t)= at^2 + bt + (-\frac{7}{3}a -  \frac{1}{2}b)$$
Thus $$N\big(T\big)= \big\{p(t) \mid p(t) = at^2 + bt + (-\frac{7}{3}a -  \frac{1}{2}b) \big\} $$
Notice that $$p(t) = at^2 + bt + (-\frac{7}{3}a -  \frac{1}{2}b)$$
$$= a(t^2 - \frac{7}{3}) + b(t - \frac{1}{2})$$
Since these two polynomials are of different degrees then they are linearly independent and thus they constitute a basis for the null space, that is
$$Basis = \big \{t^2 - \frac{7}{3}, t - \frac{1}{2} \big \}$$
$$\implies dim(N(T)) = 2$$
$$\implies dim(R(T)) = 3 -2 = 1$$
Since this function is not one-to-one, that is $dim(N(T)) \neq 0$, then it cannot be an isomorphism.
