Show that there is a unique polynomial $\int_{x}^{x+1} b_{n}{(t)} dt = x^n$ For each n = 1, 2, 3, . . . show that there is a unique polynomial $ b_{n}{(x)} $ that
satisfies the equation
$\int_{x}^{x+1} b_{n}{(t)} dt = x^n$
n = 1, I find $ b_{1}{(t)} = t + x_{0} $ and find $x_{0} = -\frac{1}{2}$
n = 2, $ b_{2}{(t)} = t^2 - t + \frac{1}{6} $
How  can I prove that the polynomial is unique and always satisfy the equation?

First observe that
$$
\int_x^{x + 1} t^n \,dt = \frac{(x + 1)^{n + 1} - x^n}{n+1} = \frac{1}{n + 1} \sum_{k = 0}^n {n + 1 \choose k} x^k.
$$
Therefore
$$\int_x^{x+1} t^n \,dt = x^n + \text{lower order terms}.$$
Therefore the matrix which writes $\{\int_{x}^{x+1} t^n \,dt : n = 0, 1, 2, 3\dots \}$ in terms of $\{1,x,x^2,x^3,\dots\}$ is triangular with ones down the diagonal. In particular, it is invertible. Because $\{1,x,x^2,x^3,\dots\}$ is a basis for the ring of polynomials, this means that $\{\int_{x}^{x+1} t^n \,dt : n = 0, 1, 2, 3\dots \}$ is also a basis. Therefore every polynomial in $x$ can be written uniquely as a linear combination
$$
\begin{align*}
p(x) &= a_n \int_{x}^{x + 1} t^n \,dt + a_{n - 1}\int_x^{x+1} t^{n - 1} \,dt + \dots + a_1 \int_x^{x+1} t \,dt + a_0 \int_x^{x+1} 1\, dt \\
&= \int_x^{x+1} a_nt^n + a_{n-1}t^{n-1} + \dots + a_1t + a_0 \, dt.
\end{align*}
$$
