# Prove that I(n+2)+2010^2*I(n)=1/(n+1) if I(n) is the definite integral of y=x^n/(x^2+2010^2) for x=0 to x=1 .

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### 3 Answers

I(n) = intg x^n / (x^2 + 2010^2)

I(n+1) = intg (x^n+1) / (x^2 + 2010^2)

I(n+2) = intg (x^(n+2) / (x^2+ 2010^2)

Let us integrate:

I(n+2) = intg (x^(n+2) /(x^2 + 2010^2) dx

= intg (x^n)(x^2)/(x^2 + 2010^2) dx

= intg (x^n) - intg ((2010^2*x^n/(x^2+2010^2) dx

= intg x^n - 2010^2 intg (x^n/(x^2 + 2010^2 ) dx

= intg x^n - 2010^2 * I(n)

= 1/(n+1) - 2010^2 * I(n)

==> I(n+2) = 1/(n+1) - 2010^2 * I(n)

**==> I(n+2) + 2010^2* I(n) = 1/(n+1) **

I(n) = Integral { x^n /(x^2+2010^22) } dx from x = 0 to x= 1.

Therefore we get :

A...I(n+2) = Integral x^(n+2) dx/(x^2+2010^2) ..x = 0 to x= 1.

I(n) = Integral x^n dx /(x^2+2010^2) for x = 0 to x = 1.

B....2010^2 *I(n) = Integral 2010^2 * x^n/(x^2+2010^2) for x= 0 to x =1.

Therefore adding A and B we get:

I(n+2) +2010^2I(n) = Integral (x^n+2)+ 2010^2*x^n)dx)/(x^2+2010^2) for x= 0 to x = 1.

I(n+2) +2010^2I(n) = Integral x^n (x^2+2010^2)/(x^2+2010)^2 forx = 0 to x= 1.

I(n+2) +2010^2I(n) = Integral x^n for x= 0 to x = 1.

I(n+2)+2010^2 I(n) = {(1/(n+1)) x^(n+1) at x= 1} - {(1/(n+1)) x^(n+1) at x= 0} = {1/(n+1)}(1 - 0}

Therefore I(n+2)- I(n) = {1/(n+1)}.

We'll write the definite integral of I(n):

I(n) = Int x^n dx/(x^2+2010^2)

We'll multiply I(n) by 2010^2:

2010^2I(n) = Int 2010^2*x^n dx/(x^2+2010^2) (1)

We'll write the definite integral of I(n+2):

I(n+2) = Int x^(n+2) dx/(x^2+2010^2) (2)

We'll apply the property of integral to be additive:

(1) + (2) = Int [x^(n+2) + 2010^2*x^n]dx/(x^2+2010^2)

Well factorize by x^n:

Int [x^n(x^2 + 2010^2)]dx/(x^2+2010^2)

We'll simplify by (x^2+2010^2) and we'll get:

Int x^n dx = x^(n+1)/(n+1)

For x = 0 => 0^(n+1)/(n+1) = 0

For x = 1 => 1/(n+1)

We'll apply Leibniz-Newton:

Int x^n dx = F(1) - F(0)

Int x^n dx = 1/(n+1)

**I(n+2) + 2010^2I(n) = Int x^n dx = 1/(n+1) q.e.d.**