Let I_n(x) = ₀^x t^n e^ -t^2 , dt for n 1 . Then the expression 2x I₅(x) + I'₅(x) is equal to

Step-by-step solution

Given, I_n(x) = ₀^x t^n e^ -t^2 , dt We can write this as I_n(x) = ₀^x t^ n-1 (t e^ -t^2 ) , dt . Applying integration by parts by taking u = t^ n-1 and dv = t e^ -t^2 dt , we get: I_n(x) = [ t^ n-1 (- 1 2 e^ -t^2 ) ]₀^x - ₀^x (n-1)t^ n-2 (- 1 2 e^ -t^2 ) , dt I_n(x) = - 1 2 x^ n-1 e^ -x^2 + n-1 2 ₀^x t^ n-2 e^ -t^2 , dt I_n(x) = - 1 2 x^ n-1 e^ -x^2 + n-1 2 I_ n-2 (x) By Leibniz rule, differentiating I_n(x) with respect to x gives: I'_n(x) = x^n e^ -x^2 Substituting x^ n-1 e^ -x^2 = 1 x I'_n(x) into our equation: