integration of polynomial


Theorem.

For all nonnegative integers n,

xn𝑑x=1n+1xn+1+C.
Proof.

It will first be proven that, for any nonnegative integer n and any a,

0axn𝑑x=1n+1an+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:

0axn𝑑x =limtk=1t(akt)n(at)
=an+1limt1tn+1k=1tkn
=an+1limt1tn+1l=1n+1(n+1r)Bn+1-ln+1(t+1)l by this theorem (http://planetmath.org/SumOfKthPowersOfTheFirstNPositiveIntegers),
=an+1limt1tn+1(n+1n+1)Bn+1-(n+1)n+1(t+1)n+1
=B0n+1an+1limt(t+1t)n+1
=1n+1an+1

Thus, if a,b, then abxn𝑑x=0bxn𝑑x-0axn𝑑x=1n+1bn+1-1n+1an+1.

It follows that xn𝑑x=1n+1xn+C. ∎

Title integration of polynomial
Canonical name IntegrationOfPolynomial
Date of creation 2013-03-22 15:57:29
Last modified on 2013-03-22 15:57:29
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 30
Author Wkbj79 (1863)
Entry type Theorem
Classification msc 26A42