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integration of polynomial
Theorem For all nonnegative integers $n$ ,$$ \int x^n \, dx = \frac{1}{n+1}x^{n+1}+C.$$
Proof. It will first be proven that, for any nonnegative integer $n$ and any $a \in \mathbb{R}$ ,$$ \int\limits_0^a x^n \, dx = \frac{1}{n+1}a^{n+1}.$$
If $a=0$ , the above statement is obvious. If $a>0$ , the following computation uses the right hand rule for computing the integral; if $a<0$ , the following computation uses the left hand rule for computing the integral:
| $\displaystyle \int\limits_0^a x^n \, dx$ | $\displaystyle =\lim_{t \to \infty} \sum_{k=1}^t \left( \frac{ak}{t} \right)^n \left( \frac{a}{t} \right)$ |
| $\displaystyle =a^{n+1} \lim_{t \to \infty} \frac{1}{t^{n+1}} \sum_{k=1}^t k^n$ | |
| $\displaystyle =a^{n+1} \lim_{t \to \infty} \frac{1}{t^{n+1}} \sum_{l=1}^{n+1} {n+1 \choose r} \frac{B_{n+1-l}}{n+1}(t+1)^l$ by this theorem, | |
| $\displaystyle =a^{n+1} \lim_{t \to \infty} \frac{1}{t^{n+1}} {n+1 \choose n+1} \frac{B_{n+1-(n+1)}}{n+1} (t+1)^{n+1}$ | |
| $\displaystyle =\frac{B_0}{n+1} a^{n+1} \lim_{t \to \infty} \left( \frac{t+1}{t} \right)^{n+1}$ | |
| $\displaystyle =\frac{1}{n+1}a^{n+1}$ |
Thus, if $a,b \in \mathbb{R}$ , then $\displaystyle \int\limits_a^b x^n \, dx = \int\limits_0^b x^n \, dx - \int\limits_0^a x^n \, dx = \frac{1}{n+1}b^{n+1}-\frac{1}{n+1}a^{n+1}.$
It follows that $\displaystyle \int x^n \, dx = \frac{1}{n+1}x^n+C$ . 
integration of polynomial is owned by Warren Buck.
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