The integralI

3

=

1

0

x

3

e

x

dx

can be evaluated by three successive applications of integration by parts. An elegant way to perform successive applications of integration by parts is to use a reduction formula.

The fist step, much like recusion or induction, is to have a base case. This is typically a simple integral such as:

Consider the denite integral 1 In = /xne_x dx. 0 The integral I3 = [01 x3e_xdx can be evaluated by three successive applications of integration by parts. An elegant way to perform successive applications of integration by parts is to use a reduction formula. The fist step, much like recusion or induction, is to have a base case. This is typically a simple integral such as: 1 10=/e'xdx=l n. 0 The following reduction formula is a compact way of writing the result of integration by parts for the integral In: 1n=nI _1%forn21 Using this reduction formula 1 III '11=10-z= :ll ~12=211-%= IE] -Is=312%= IE1- AI_-II-,__,-I \\I-,, A-_,, I!I,_ 1- 4-4\"- u._ A_J,,_1!-,_ z_4,a_,,|- ,,-,,A--I1