int. (x^n)(e^x) dx patterns

[zketrouble]

so I just introduced myself to the concept of integration by parts a few days ago, and today for practice I decided to try to redo one of the examples given in a youtube video without the assistance of youtube. That problem was int. (x^2)(e^x) dx, for which the answer is (e^x)(x^2 -2x + 2). So then I tried to integrate (x^3)(e^x), (x^4)(e^x), and so on, a drew the results in a table:

y = int. (x^n)(e^x)

n=0; y = e^x
n=1; y = e^x(x-1)
n=2; y = e^x(x^2 - 2x + 2)
n=3; y = e^x(x^3 - 3x^2 + 6x -6)
n=4; y = e^x(x^4 - 4x^3 + 12x^2 - 24x + 24)
n=5; y = e^x(x^5 - 5x^4 + 20x^3 - 60x^2 + 120x -120)

I noticed a pattern, and I'm curious as to how I would formally prove this pattern:

(x^n)(e^x) dx =
(e^x){x^n - nx^(n-1) + n(n-1)x^(n-2) - n(n-1)(n-2)x^(n-3) + n(n-1)(n-2)(n-3)x^(n-4)...-+-+-+...}

It is very likely that the procedures necessary to prove this are way beyond my current calculus skills, but I'm curious anyway to see what the folks here at PF can come up with to prove/disprove my assumption that the above statement is true.

[dextercioby]

It's not calculus needed to prove the general 'n' case, but mathematical induction.

Yay, 10^2^2 posts :biggrin:

[zketrouble]

silly me...forgot all of the + c every time. :p

[Goongyae]

>n=5; y = e^x(x^5 - 5x^4 + 20x^3 - 60x^2 + 120x -120)
>I noticed a pattern, and I'm curious as to how I would formally prove this pattern:

You don't need induction to prove your pattern. If you take the derivative of one of your examples

d/dx [e^x(x^5 - 5x^4 + 20x^3 - 60x^2 + 120x -120)] =
e^x(x^5 - 5x^4 + 20x^3 - 60x^2 + 120x -120)
+e^x(5x^4 - 20x^3 + 60x^2 -120x + 120) = e^x x^5

you can see that every term but the first vanishes. Since the expression produces the derivative you wanted (e^x x^5), it must be the correct antiderivative (though any constant can be added of course). Just do the same thing with your formula and you've proven it.

[fzero]

An advanced concept that is very useful here is that of a generating function. Consider

F(a) = \int e^{ax} dx= \frac{e^{ax}}{a} + c.

We can compute

\frac{d^nF(a)}{da^n} = \int x^n e^{ax} dx,

so that

I_n = \int x^n e^x dx = \left. \frac{d^nF(a)}{da^n}\right|_{a=1}.

Therefore we can generate any of the I_n just by computing derivatives of F(a). These derivatives are just the same derivatives that you compute when integrating by parts.

[atomthick]

your result is correct

let In be your integral; if you integrate by parts In = xnex - nIn-1