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

14.2 Integration by parts 459
Usingthe integration by parts formula fordefinite integrals we have
∫ 2
∫ 2
xe x dx = [xe x ] 2 0 − e x·1dx
0
0
= 2e 2 −[e x ] 2 0
=2e 2 −[e 2 −1]
=e 2 +1
Sometimes integration by parts needs tobe used twice,as the next example illustrates.
Example14.3 Evaluate
∫ 2
0
x 2 e x dx
Solution We let
Then
u=x 2
and
dv
dx = ex
du
=2x and v=ex
dx
Usingthe integration by parts formula wehave
Now
∫ 2
0
∫ 2
0
∫ 2
0
∫ 2
x 2 e x dx = [x 2 e x ] 2 0 − 2xe x dx
=4e 2 −2
0
∫ 2
0
xe x dx
xe x dx has been evaluated usingintegration by parts inExample 14.2. So
x 2 e x dx = 4e 2 −2[e 2 +1] = 2e 2 −2 = 12.78
The next example illustrates a case in which the integral to be found reappears after
repeated application ofintegration by parts.
Example14.4 Find
∫
e t sintdt