Let I_n = ₀^1 x^n 1-x^2 , dx for n = 0, 1, 2, . If 102 101 100 I₁₀₀ I₉₉ = k , then the value of the integer k is equal to _______.

Step-by-step solution

Let I_n = ₀^1 x^n 1-x^2 , dx . For n 2 , use integration by parts with u = x^ n-1 and dv = x 1-x^2 , dx . Then du = (n-1)x^ n-2 , dx and v = - 1 3 (1-x^2)^ 3/2 . I_n = [ -x^ n-1 1 3 (1-x^2)^ 3/2 ]₀^1 + n-1 3 ₀^1 x^ n-2 (1-x^2)^ 3/2 , dx The boundary term is 0 . I_n = n-1 3 ₀^1 x^ n-2 (1-x^2) 1-x^2 , dx I_n = n-1 3 ( ₀^1 x^ n-2 1-x^2 , dx - ₀^1 x^n 1-x^2 , dx ) I_n = n-1 3 (I_ n-2 - I_n) Rearranging gives: 3 I_n = (n-1) I_ n-2 - (n-1) I_n (n+2) I_n = (n-1) I_ n-2 Multiply both sides by (n+1)n I_ n-1 : (n+2)(n+1)n I_