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 (http://planetmath.org/RiemannIntegral); 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 (http://planetmath.org/SumOfKthPowersOfTheFirstNPositiveIntegers),
$\displaystyle=a^{n+1}\lim_{t\to\infty}\frac{1}{t^{n+1}}{n+1\choose n+1}\frac{B% _{n+1-(n+1)}}{n+1}(t+1)^{n+1}$
$\displaystyle=\frac{B_{0}}{n+1}a^{n+1}\lim_{t\to\infty}\left(\frac{t+1}{t}% \right)^{n+1}$
$\displaystyle=\frac{1}{n+1}a^{n+1}$

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 $\displaystyle\int x^{n}\,dx=\frac{1}{n+1}x^{n}+C$. ∎

Title integration of polynomial IntegrationOfPolynomial 2013-03-22 15:57:29 2013-03-22 15:57:29 Wkbj79 (1863) Wkbj79 (1863) 30 Wkbj79 (1863) Theorem msc 26A42