Chabot Mathematics
§6.1 Integ
by PARTS
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
§6.1 Learning Goals
 Use integration by parts to find integrals
and solve applied problems
 Examine and use a table of integrals
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Integration by Parts
 Integration by Parts is based on the
Derivative Product Rule
d
dv
du
u x   vx   u x   vx 
dx
dx
dx
 Isolate u·[dv/dx] by Algebra
dv d
du
u  x   u  x   v x   v x 
dx dx
dx
 Now multiply both sides by dx
dv d
du 

u x  dx  dx u x   vx   vx  dx   dx
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Integration by Parts
 Now Integrate with respect to x










u
x
dv

1

d
u
x

v
x

v
x
du



 ux dv  ux  vx    vx du
 Or as it is usually written
u
dv

u

v

vdu


Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
When to Use “Parts”
 Need Integrand of this form
𝑢 ∙ 𝑑𝑣 =
𝑓 𝑥
∙ 𝑔 𝑥 ∙𝑥
 ReConfigure Integrand to optimum
Condition by
• COMMUTATION (Switching) of the
Integrand Factors
• ASSOCIATION (Grouping) of the Integrand
Factor
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
“Parts” Examples
 On WhiteBoard:
MTH16 • Bruce Mayer, PE
 Evaluate
3
 x  ln x dx
y = x•ln(x)
2.5
2
1.5
1
0.5
0
MTH15 Quick Plot BlueGreenBkGnd 130911.m
 Find the FORMULA for
0.5
1
1.5
2
2.5
3
x


ln
x
dx

Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
RePeated “Parts”
 ReCall Integration by Parts Formula
u
dv

u

v

vdu


 “Expand” by Parts the R.H.S. Integral
u
dv

u

v



 pdq
u
dv

u

v

p

q

qdp


Chabot College Mathematics
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
RePeated “Parts”
 Continue the Expansion
 
u
dv

u

v

p

q

rds



u
dv

u

v

p

q

r

s

sdr


u
dv

u

v

p

q

r

s

sdr


 This Process can continue until
d”u”/dx = 0
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx

RePeated“Parts” Examples
 On WhiteBoard:
MTH16 • Bruce Mayer, PE
7
 Evaluate
x

e

dx

3
5
y = x3•ex/2
x
2
6
4
3
2
1
0
MTH15 Quick Plot BlueGreenBkGnd 130911.m
0
0.5
1
1.5
x
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
“Part” Utility
 Integration by Parts can be extremely
useful when the R.H.S. vdu integral is
easier to find than the L.H.S. udv integral




Chabot College Mathematics
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
Example  Find
x e dx
SOLUTION
Note that
can NOT be integrated
Thus ReWrite
Integral as
2
x2
3 x2




x
xe
d
x

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
“Part” Utility
 Recognize
 Then dv
 
dv  xe dx 
x2




x
xe
d
x

x2
2
u
 dv   
ux


1 x2
xe dx  v  e
2
 And du
2
dv
d
d 2
u
x
dx
dx
x2
du
 2 x  du  2 xdx
dx

 Sub u & v into “Parts” Eqn v
1 x2
1 x2
 udv  u  v   vdu   x  xe dx  x  2 e   2 e  2 x dx
2
Chabot College Mathematics
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x2
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
“Part” Utility
1 x2
1 x2
 Thus  x  xe dx  x  e   e  2 xdx
2
2
 So
1 x2
2
x2
2 1 x2
 x  xe dx  x  2 e  2 e  C
2
x2
2
 Factoring a bit


1 x2 2
x

xe
dx

e
x

1

C

2
2
Chabot College Mathematics
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x2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Integration by Parts Procedure
 “Parts” is most useful when Integrands
contain Products in which one Factor
is difficult (or impossible) to integrate
 The main step to Integration by Parts
1. Chose u & v so that f(x)dx = u·dv
•
Since finding du/dx is EASY, then u
should be difficult to Integrate
•
Since finding ∫du might be DIFFICULT,
then dv should be easy to Integrate
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Integration by Parts Procedure
2. Organize
u & dv as:
u x 

du dx

du
dv

dv  qx dx

 dv   qx dx
3. Complete the process by finding ∫v·du
 f x dx   udv uv   vdu
•
Add the const, C, at the very end
of the Process
Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Table of Integrals
 Common Integrals are often Tabulated
as is done in Tab 6.1 in the Text
 The Key to Using Integral Tables
Make the Integral-of-Interest
“Look Like” the Tabulated
Integral through the
Application of a CLEVER
SUBSTITION
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Table of Integrals
 Evaluate
25  x 2
dx
x

 SOLUTION
 The format of this integral is NOT
solvable DIRECTLY using either Parts
or Substitution
 Now use the indirect substitutions
a  25  a  5
2
u  x  du  dx
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Table of Integrals
 Making the Substitutions

25  x 2
dx 
x

a2  u2
du
u
 Consult the Text table of integrals to
find that the substituted integral is of the
form of #17 from the text; then

2
2
a2  u2
a

a

u
du  a 2  u 2  a ln
 C.
u
u
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Table of Integrals
 ReCall the
a 2  25 a  5 u  x
Substitutions
 Making the Substitutions in the solution

2
2
a2  u2
a

a

u
du  a 2  u 2  a ln
 C.
u
u

2
25  x 2
5

25

x
dx  25  x 2  5 ln
 C.
x
x
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
WhiteBoard Work
 Problems From §6.1
• P66 → Centroid of Area
– Covered in Detail in ENGR36
MTH16 • Bruce Mayer, PE
2
y = ln(x)
1.5
1
0.5
0
MTH15 Quick Plot BlueGreenBkGnd 130911.m
0
Chabot College Mathematics
19
1
2
3
4
x
5
6
7
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
All Done for Today
Parts
is
Parts
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
P6.1-66 MATLAB Plot
% Bruce Mayer, PE
% MTH-15 •10Jan14
% MTH15_Quick_Plot_BlueGreenBkGnd_130911.m
%
clear; clc; clf; % clf clears figure window
%
% The Domain Limits
xmin = 1; xmax = exp(1)^2;
% The FUNCTION **************************************
x = linspace(xmin,xmax,10000); y = log(x);
% ***************************************************
% the Plotting Range = 1.05*FcnRange
% ymin = min(y); ymax = max(y); % the Range Limits
ymin = -0.25; ymax =2.25;
R = ymax - ymin; ymid = (ymax + ymin)/2;
ypmin = ymid - 1.025*R/2; ypmax = ymid + 1.025*R/2
xmin = 0; xmax = 8;
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ypmin*1.05 ypmax*1.05];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([1 1 1]);
area(x,y, 'LineWidth', 4, 'FaceColor',[1 1 .69]),grid, axis([xmin xmax ypmin
ypmax]),...
xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = ln(x)'),...
title(['\fontsize{16}MTH16 • Bruce Mayer, PE',]),...
annotation('textbox',[.53 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor',
'none', 'String', 'MTH15 Quick Plot BlueGreenBkGnd 130911.m','FontSize',7)
hold on
plot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)
hold off
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx