082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

Usingintegration by parts we have
∫ 2
0
Writing
I n
=
wesee that
Hence
I n−1
=
∫ 2
x n e x dx = [x n e x ] 2 0 − nx n−1 e x dx
∫ 2
0
∫ 2
=2 n e 2 −n
x n e x dx
0
x n−1 e x dx
0
∫ 2
0
x n−1 e x dx
14.2 Integration by parts 461
I n
= 2 n e 2 −nI n−1
(14.3)
Equation (14.3) iscalled areduction formula.
We have already evaluatedI 1
, thatis
I 1
=e 2 +1
∫ 2
Usingthe reduction formula withn = 2 gives
∫ 2
0
x 2 e x dx =I 2
=2 2 e 2 −2I 1
= 4e 2 −2(e 2 +1)
=2e 2 −2
Note that thisisinagreement with Example 14.3.
Withn = 3 the reduction formulayields
∫ 2
0
x 3 e x dx =I 3
=2 3 e 2 −3I 2
Withn = 4 we have
∫ 2
0
= 8e 2 −3(2e 2 −2)
=2e 2 +6
x 4 e x dx =I 4
=2 4 e 2 −4I 3
Withn = 5 we have
∫ 2
= 16e 2 −4(2e 2 +6)
=8e 2 −24
0
xe x dx inExample 14.2, and found
0
x 5 e x dx =I 5
=2 5 e 2 −5I 4
= 32e 2 −5(8e 2 −24)
= 120 −8e 2