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#1
11-02-2006, 01:15 PM
 DudeMan Member Join Date: Jan 2006 Location: teh Po' Studying for shts&ggles Favorite beer: Ginger Beer Posts: 11,609
Integration by parts shortcut

I need to integrate x^4*e^(-2x)....any suggestions?
#2
11-02-2006, 01:31 PM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

Quote:
 Originally Posted by GraffixMan I need to integrate x^4*e^(-2x)....any suggestions?
Tabular integration by parts. Here's how it works: Write down the derivatives of $x^4$ alongside the antiderivatives of $e^{-2x}$:

$x^4 \qquad e^{-2x}$

$4x^3 \qquad -\frac{1}{2}e^{-2x}$

$12x^2 \qquad \frac{1}{4}e^{-2x}$

$24x \qquad -\frac{1}{8}e^{-2x}$

$24 \qquad \frac{1}{16}e^{-2x}$

$0 \qquad -\frac{1}{32}e^{-2x}$

Now form the alternating sum of the products of the k(th) entry in the first column with the (k+1)th entry in the second column until such products are not possible:

$x^4 (-\frac{1}{2}e^{-2x}) - 4x^3 (\frac{1}{4}e^{-2x}) + 12x^2 (-\frac{1}{8}e^{-2x}) - 24x (\frac{1}{16}e^{-2x}) + 24 (-\frac{1}{32}e^{-2x})$

which in turn simplifies to

$-e^{-2x}\left(\frac{1}{2}x^4 + x^3 + \frac{3}{2}x^2 + \frac{3}{2}x + \frac{3}{4}\right)$.

This is the required antiderivative.
#3
11-02-2006, 01:32 PM
 SirVLCIV Member SOA Join Date: Feb 2006 Posts: 45,565

What he said.
#4
11-02-2006, 01:37 PM
 Kabaka Member SOA Join Date: Aug 2006 Location: O Canada Studying for NOTHING! :) Favorite beer: Root Posts: 2,186

Neato. Good tip.
#5
11-02-2006, 01:42 PM
 mlschop Member SOA Join Date: Sep 2005 Posts: 35,808

This should be a sticky! I seriously have never heard this trick...ever.

Does it work with anything (i.e. instead of exponentials, would it work with trig functions - as long as one terminations with increasing derivatives?)
#6
11-02-2006, 02:03 PM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

Tabular integration is sometimes taught at the AP Calculus BC level, and is mentioned in the Wikipedia entry on integration by parts. It is merely a bookkeeping method for a recursive integration. (Bookkeeping, by the way, is one of the few words which have three consecutive double letter pairs.) Thus the method is valid for any integrand that is the product of two functions, one of which has an n(th) derivative that is zero, and the other has n successive antiderivatives.

Incidentally, this is a method by which one can easily prove the formula for the CDF of a gamma distribution.
#7
11-02-2006, 02:12 PM
 SirVLCIV Member SOA Join Date: Feb 2006 Posts: 45,565

Quote:
 Originally Posted by atomic Tabular integration is sometimes taught at the AP Calculus BC level, and is mentioned in the Wikipedia entry on integration by parts. It is merely a bookkeeping method for a recursive integration. (Bookkeeping, by the way, is one of the few words which have three consecutive double letter pairs.) Thus the method is valid for any integrand that is the product of two functions, one of which has an n(th) derivative that is zero, and the other has n successive antiderivatives. Incidentally, this is a method by which one can easily prove the formula for the CDF of a gamma distribution.
IIRC (I'd need to do it again), a buddy and I both (independently) solved a problem he had in a "book of problems" similarly. However, both multiples were recursive (one was trig and the other exponential IIRC). I'll have to ask him for the problem again, but I believe he intended to generalize it and translate it into a useful Laplace transform.

(it's probably been done before, but we're undergrads, and it's good practice ).
#8
11-02-2006, 02:25 PM
 Kazodev Member SOA Join Date: May 2004 Posts: 3,393

Doesn't Laplace have bounds from 0 to infnity so lots of things cancel?
#9
11-02-2006, 03:15 PM
 J_C Member CAS Join Date: May 2006 Posts: 2,834

tic-tac-toe method
#10
11-02-2006, 03:41 PM
 Qisoneminusp Member Join Date: Jun 2006 Location: in the saddle Studying for Run DMaC Posts: 2,909

Isnt that tabular method just a more organized way of doing nested uv-int(vdu) over and over again until du is 1?

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