integration of polynomial


Theorem.

For all nonnegative integers n,

∫xn⁢𝑑x=1n+1⁢xn+1+C.
Proof.

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

∫0axn⁢𝑑x=1n+1⁢an+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:

∫0axn⁢𝑑x =limt→∞⁡∑k=1t(a⁢kt)n⁢(at)
=an+1⁢limt→∞⁡1tn+1⁢∑k=1tkn
=an+1⁢limt→∞⁡1tn+1⁢∑l=1n+1(n+1r)⁢Bn+1-ln+1⁢(t+1)l by this theorem (http://planetmath.org/SumOfKthPowersOfTheFirstNPositiveIntegers),
=an+1⁢limt→∞⁡1tn+1⁢(n+1n+1)⁢Bn+1-(n+1)n+1⁢(t+1)n+1
=B0n+1⁢an+1⁢limt→∞⁡(t+1t)n+1
=1n+1⁢an+1

Thus, if a,b∈ℝ, then ∫abxn⁢𝑑x=∫0bxn⁢𝑑x-∫0axn⁢𝑑x=1n+1⁢bn+1-1n+1⁢an+1.

It follows that ∫xn⁢𝑑x=1n+1⁢xn+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