## Wednesday, January 5, 2022

### 12 Days of Christmas Integrals: ∫ x^n ∙ e^x dx; n is a positive integer

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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