A problem related to coefficient of $\int{(t-a)^n}dt$ Let's make some antiderivative of $(t-a)^n$.
When $n=1$, we get $\frac{1}{2}t^2-at$ for the antiderivative (ignoring some constant). Now when we put $t=a$, we get $-\frac{1}{2}a^2$ and $-\frac{1}{2}$ is the coefficient.
When $n=3$, we get $\frac{1}{4}t^4-at^3+\frac{3}{2}a^2t^2-a^4t$ for the antiderivative and we get $-\frac{1}{4}$ for the coefficient by putting $t=a$.
When $n=5$, in the same manner, we get $-\frac{1}{6}$ for the coefficient.
The case when exponent $n$ is even, it is straightforward to know that the coefficient always comes out as $\frac{1}{n+1}$. However I'm stuck with finding a proof for when $n$ is odd.
Any help would be grateful, thanks.
Edit: I've already gotten a great answer however my original purpose of posing this question was to approach this question in regard of the pattern of binomial coefficients. If there is someone who knows some proofs that looks somewhat more 'discrete', I would be grateful to know any of them. Thanks.
Edit: My question can be substituted as follows.
Let's make an array of coefficients of $(a-b)^n$ as Pascal's Triangle.
Then we can see the following.
\begin{array}{c}
+1 \\
+1 \quad -1 \\
+1 \quad -2 \quad +1 \\
+1 \quad -3 \quad +3 \quad -1 \\
+1 \quad -4 \quad +6 \quad -4 \quad +1 \\
+1 \quad -5 \quad +10 \quad -10 \quad +5 \quad -1 \\
+1 \quad -6 \quad +15 \quad -20 \quad +15 \quad -6 \quad +1 \\
+1 \quad -7 \quad +21 \quad -35 \quad +35 \quad -21 \quad +7 \quad -1 \\
\end{array}
Now, let's multiply each coefficient by $\frac{1}{n+1}$, $\frac{1}{n}$, $\frac{1}{n-1}$, $\frac{1}{n-2}$ $\cdots$, $\frac{1}{1}$ as the following.
\begin{array}{c}
\frac{+1}{1} \\
\frac{+1}{2} \quad \frac{-1}{1} \\
\frac{+1}{3} \quad \frac{-2}{2} \quad \frac{1}{1} \\
\frac{+1}{4} \quad \frac{-3}{3} \quad \frac{+3}{2} \quad \frac{-1}{1} \\
\frac{+1}{5} \quad \frac{-4}{4} \quad \frac{+6}{3} \quad \frac{-4}{2} \quad \frac{1}{1} \\
\frac{+1}{6} \quad \frac{-5}{5} \quad \frac{+10}{4} \quad \frac{-10}{3} \quad \frac{+5}{2} \quad \frac{-1}{1} \\
\frac{+1}{7} \quad \frac{-6}{6} \quad \frac{+15}{5} \quad \frac{-20}{4} \quad \frac{+15}{3} \quad \frac{-6}{2} \quad \frac{1}{1} \\
\frac{+1}{8} \quad \frac{-7}{7} \quad \frac{+21}{6} \quad \frac{-35}{5} \quad \frac{+35}{4} \quad \frac{-21}{3} \quad \frac{+7}{2} \quad \frac{-1}{1} \\
\end{array}
When $n$ is even, then it is straightforward that the sum of each floor equals to $\frac{1}{n+1}$.
However, when $n$ is odd, the numbers get quite messed up. So I'm stuck with deducing some 'discrete' proof that it equals to $-\frac{1}{n+1}$.
If someone knows some other approach than using the fundamental theorem of calculus it would be grateful if I could know them. Thanks.

Setting the constant equal to $0$ and taking $t=a$ means you are computing :
$$I_n = \int_0^a (t-a)^n\text dt$$
A change of variable $x = t-a$ tells us that :
$$I_n = \int_0^a(-x)^n\text dx$$
And we can compute :
$$I_n = \frac{(-1)^n a^{n+1}}{n+1}$$
In particular when $n$ is odd, you have :
$$I_n = -\frac{a^{n+1}}{n+1}$$
Another proof
\begin{align}
I_n &= \int_0^a (t-a)^n \text dt \\
&= \sum_{k=0}^n {n \choose k}\int_0^a t^{k} (-a)^{n-k}\text dt \\
&= a^{n+1}\sum_{k=0}^n {n\choose k} \frac{(-1)^{n-k}}{k+1} \\
\end{align}
Then, we use  :
$$\frac{1}{k+1}{n\choose k} = \frac{1}{n+1}{n+1\choose k+1}$$
Therefore :
\begin{align}
I_n &= \frac{a^{n+1}}{n+1} \sum_{k=0}^{n} {n+1 \choose k+1} (-1)^{n-k} \\
&=\frac{a^{n+1}}{n+1} \sum_{k=1}^{n+1} {n+1 \choose k} (-1)^{n+1-k}\\
&= \frac{a^{n+1}}{n+1}\left( \sum_{k=0}^{n+1} {n+1 \choose k} (-1)^{n+1-k} - (-1)^{n+1} \right) \\
&= \frac{a^{n+1}}{n+1}\left(- (-1)^{n+1} \right) \\
&= \frac{(-1)^n a^{n+1}}{n+1}
\end{align}
