Let $U=\{p\in P_4(\mathbb R):\int_{-1}^1 p=0\}$ Find a basis for U I am able to do the mechanics of the question that isn't an issue. I was curious about the existence of this subspace. I ask this because if you integrate a polynomial of degree 4 you obviously get a polynomial of degree 5. But how is a polynomial of degree 5 able to exist in this subspace if it is specified that the space is made up of $p\in P_4(\mathbb R)$?
 A: There's nothing that would require a polynomial of degree 5 to exist within the space. First I have the vector space $V$, consisting of $p\in P_4(\mathbb{R})$. This is all 4th-order polynomials. Then I take subspace $U$ which has the definite integral on $[-1,1]$ come out to 0. The map $f : P_4(\mathbb{R}) \to \mathbb{R}$ given by $f(p) = \int_{-1}^1 p$ is a linear transformation (basically: a matrix), and we're asking for the kernel of this transformation. And the kernel of a linear transform is always another subspace!
A: It helps me to think of the problem in terms of functions and their domains. Consider the function $L$ which sends polynomials to real numbers in the following way: $$L(p) \equiv \int_{-1}^1 p(x) \,dx.$$
So we can consider $L$ to be a function from $P_4(\mathbb{R})$ to $\mathbb{R}$, and we can consider your set to be 
$$U \equiv \{ p \in P_4(\mathbb{R}) \,:\, L(p) = 0\}.$$
Hopefully, this makes it clear that $U$ is the space of all degree-four polynomials $p$ such that $L(p)$, which is a real number, evaluates to zero.

As for where degree-five polynomials come in, you can consider $L$ to be the composition of two smaller functions:
$$\begin{align*}\mathsf{antideriv}:P_4(\mathbb{R})\rightarrow P_5(\mathbb{R})&,\qquad \mathsf{antideriv}(p) \equiv \int p(x)\,dx \\
\mathsf{evaluate}:P_5(\mathbb{R})\rightarrow \mathbb{R}&, \qquad \mathsf{evaluate}(q) \equiv q(1) - q(-1).\\\end{align*}$$
And so, if it helps to think about it this way, you can write your subset as:
$$U \equiv \left\{p \in P_4(\mathbb{R}) \,: \, \mathsf{evaluate}(\mathsf{antideriv}(p)) = 0\right\}.$$
