12 Days of Christmas Integrals: ∫ x^n ∙ e^x dx; n is a positive integer
On the Twelfth day of Christmas Integrals, the integral featured today is...
∫ x^n ∙ e^x dx; n is a positive integer
n = 1, 2, 3, 4, ...
Each term is found by repeating the Integration by Parts method. Today, we will find a (somewhat) closed formula that applies to any positive integer n.
∫ x^n ∙ e^x dx
u = x^n
du = n ∙ x^(n-1) dx
dv = e^x dx
v = e^x
= x^n ∙ e^x - ∫ n ∙ x^(n-1) ∙ e^x dx
u = n ∙ x^(n-1)
du = n ∙ (n-1) ∙ x^(n-2) dx
dv = e^x dx
v = e^x
= x^n ∙ e^x - n ∙ x^(n-1) ∙ e^x + ∫ n ∙ (n-1) ∙ x^(n-2) ∙ e^x dx
u = n ∙ x^(n-1)
du = n ∙ (n-1) ∙ x^(n-2) dx
dv = e^x dx
v = e^x
and so on until...
u = n ∙ (n - 1) ∙ ... ∙ 2 ∙ x
du = n! dx
dv = e^x dx
v = e^x
with the closed form:
= e^x ∙ [x^n - n ∙ x^(n-1) + n ∙ (n-1) ∙ x^(n-2) - n ∙ (n-1) ∙ (n-2) ∙ x^(n-3) + ...
+ (-1)^(n mod 2) ∙ n!]
Examples:
n = 2
∫ x^2 ∙ e^x dx = e^x ∙ (x^2 - 2 ∙ x + 2)
n = 3
∫ x^3 ∙ e^x dx = e^x ∙ (x^3 - 3 ∙ x^2 + 6 ∙ x - 6)
n = 4
∫ x^4 ∙ e^x dx = e^x ∙ (x^4 - 4 ∙ x^3 + 12 ∙ x^2 - 24 ∙ x + 24)
What do you think of the integral series? Is this worth a sequel?
Eddie
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