Chabot Mathematics
§5.4 Definite
Integral Apps
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Review §
5.3
 Any QUESTIONS About
• §5.3 → Fundamental Theorem and
Definite Integration
 Any QUESTIONS
About
HomeWork
• §5.3 → HW-24
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
§5.4 Learning Goals
 Explore a general procedure for using
definite integration in applications
 Find area between two curves, and use
it to compute net excess profit and
distribution of wealth (Lorenz curves)
 Derive and apply a formula for the
average value of a function
 Interpret average value in terms of rate
and area
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Need for Strip-Like Integration
 Strip Integration
•
•
Chabot College Mathematics
4
Very often, the function
f(x) to differentiate, or the
integrand to integrate, is
TOO COMPLEX to yield
exact analytical solutions.
In most cases in
engineering or science
testing, the function f(x) is
only available in a
TABULATED form with
values known only at
DISCRETE POINTS
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Strip Integration
 Game Plan: Divide
Unknown Area into
Strips (or boxes),
and Add Up
 To Improve
Accuracy the
 TOP of the Strip
can Be
• Slanted Lines
– Trapezoidal Rule
• Parabolas
– Simpson’s Rule
• Higher Order
PolyNomials
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Strip Integration
 Game Plan: Divide
Unknown Area into
Strips (or boxes),
and Add Up
 To Improve Accuracy
1. Increase the
Number of strips;
i.e., use smaller ∆x
2. Modify Strip-Tops
–
–
–
Slanted Lines (used
most often)
Parabolas
High-Order
Polynomials
 Hi-No. of Flat-Strips
Works Fine.
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  NONconstant ∆x
 Pacific Steel Casting Company (PSC) in
Berkeley CA, uses huge amounts of
electricity during the metal-melting process.
 The PSC Materials
Engineer measures the
power, P, of a certain
melting furnace over
340 minutes as shown
in the table at right.
See Data Plot at Right.
Furnace Power Consumption
250
Power Consumption, P (kW)
200
150
100
50
0
Chabot College Mathematics
7
0
50
100
150
200
time, t (min)
250
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
300
350
Example  NONconstant ∆x
 The T-table at Right displays
the Data Collected by the
PSC Materials Engineer
 Recall from Physics that Energy
(or Heat), Q, is the time-integral
of the Power.
 Use Strip-Integration to find the
Total Energy in MJ expended by
The Furnace during this process
run
Chabot College Mathematics
8
Time
(min)
0
24
45
74
90
118
134
169
180
218
229
265
287
340
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Power
(kW)
47
107
104
146
126
178
147
211
151
233
184
222
180
247
Example  NONconstant ∆x
 GamePlan for Strip Integration
 Use a Forward Difference
approach
• ∆tn = tn+1 − tn
– Example: ∆t6 = t7 − t6 = 134 − 118 =
16min → 16min·(60sec/min) = 960sec
• Over this ∆t assume the P(t) is
constant at Pavg,n =(Pn+1 + Pn)/2
– Example: Pavg,6 = (P7 + P6)/2 =
(147+178)/2 = 162.5 kW = 162.5 kJ/sec
Chabot College Mathematics
9
Time
(min)
0
24
45
74
90
118
134
169
180
218
229
265
287
340
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Power
(kW)
47
107
104
146
126
178
147
211
151
233
184
222
180
247
Example  NONconstant ∆x
 The
GamePlan
Graphically
225
x 9
Bruce May er, PE • 25Jul13
200
175
x4
150
P (kW)
• Note the
Variable
Width, ∆x,
of the Strip
Tops
MTH15 • Variable-Width Strip-Integration
125
100
75
50
25
0
0
50
100
150
200
250
t (minutes)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
300
350
Chabot College Mathematics
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 25Jul13
% XY_Area_fcn_Graph_6x6_BlueGreen_BkGnd_Template_1306.m
%
clear; clc; clf; % clf is clear figure
%
% The FUNCTION
xmin = 0; xmax = 350; ymin = 0; ymax = 225;
x = [0 24 24 45 45 74 74 90 90 118 118 134 134 169 169 180 180 218
218 229 229 265 265 287 287 340]
y = [77 77 105.5
105.5
125 125 136 136 152 152 162.5
162.5
179 179
181 181 192 192 208.5
208.5
203 203 201 201 213.5
213.5]
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
% Now make AREA Plot
area(x,y,'FaceColor',[1 0.6 1],'LineWidth', 3),axis([xmin xmax ymin
ymax]),...
grid, xlabel('\fontsize{14}t (minutes)'), ylabel('\fontsize{14}P
(kW)'),...
title(['\fontsize{16}MTH15 • Variable-Width Strip-Integration',]),...
annotation('textbox',[.15 .82 .0 .1], 'FitBoxToText', 'on', 'EdgeColor',
'none', 'String', 'Bruce Mayer, PE • 25Jul13','FontSize',7)
set(gca,'XTick',[xmin:50:xmax]); set(gca,'YTick',[ymin:25:ymax])
set(gca,'Layer','top')
Example  NONconstant ∆x
n Time, t Power ∆t = 60*(tn+1-tn) Pavg=(Pn+1−Pn)/2 ∆Q= Pavg*∆t
(cnt) (min) (kW)
(Sec)
(kW)
(kJ)
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
0
47
24
107
45
74
110880
1260
105.5
132930
1740
125
217500
960
136
130560
1680
152
255360
960
162.5
156000
2100
179
375900
660
181
119460
2280
192
437760
660
208.5
137610
2160
203
438480
1320
201
265320
3180
213.5
678930
146
126
118
178
134
147
169
211
218
77
104
90
180
1440
151
233
229
184
265
222
287
180
340
247
Total Energy in MJ = (∑∆Q)/1000 =
Chabot College Mathematics
12
 The NONconstant
Strip-Width Integration
is conveniently done
in an Excel
SpreadSheet
 The 13 ∆Q strips Add
up to 3456.69
MegaJoules of Total
Energy Expended
3456.69
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Area Between Two Curves
 Let f and g be continuous functions, the
area bounded above by y = f (x) and
below by y = g(x) on [a, b] is
  f ( x)  g ( x) dx
b
a
y
y  f ( x)
• Provided that
f ( x)  g ( x) on  a, b.
R
• The Areal Difference
Region, R, Graphically
Chabot College Mathematics
13
a
y  g ( x)
b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
x
Example Area Between Curves
f  x   11e  x 6  9
and
2
8 x  5
g x   
 10
25
ylo = (-8/25)*(x-5)2+10 • yhi = 11e-x/6+9
 Find the area between
functions f & g over
the interval x = [0,10]
MTH15 • Area Between Curves
 The Graphs
f x   11e  9
of f & g
20
16
14
12
10
8
14
8 x  5
g x   
 10
25
2
6
4
2
0
Chabot College Mathematics
x 6
18
Bruce May er, PE • 25Jul13
0
2
4
6
x
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
10
Example Area Between Curves
 The process Graphically
MTH15 • Area Between Curves
20
20
 11e
18
10
x 6
0
16
−

9 x
18
10

0
16
2

8 x  5 
10 
dx

25 

=
12
12
10
10
8
8
6
6
4
4
4
2
2
2
10
8
6
Bruce May er, PE • 25Jul13
0
0
2
Bruce May er, PE • 25Jul13
4
6
x
8
Chabot College Mathematics
15
0
16
14
12
  f x  g xdx
10
18
14
ylo = (-8/25)*(x-5)2+10 • yhi = 11e-x/6+9
14
20
10
0
0
2
4
6
x
Bruce May er, PE • 25Jul13
8
10
0
0
2
4
6
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
8
10
Example Area Between Curves
 Do the Math →   f x  g xdx
10
0
f x  
g x  
f x   g x  
  f x   g x dx 
  f x  g xdx 
10
0
≈ 70.20
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example Area Between Curves
MTH15 • Area Between Curves
20
ylo = (-8/25)*(x-5)2+10 • yhi = 11e-x/6+9
 Thus
Ans
18
16
14
A = 70.200
12
10
8
6
4
2
0
Bruce May er, PE • 25Jul13
0
2
4
6
8
x
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
10
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 25Jun13
%
clear; clc; clf; % clf clears figure window
%
% The Limits
xmin = 0; xmax = 10;
ymin = 0; ymax = 20;
% The FUNCTION
x = linspace(xmin,xmax,500); y1 = (-8/25)*(x-5).^2 + 10; y2 = 11*exp(-x/6)+9;
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
subplot(1,3,2)
area(x,y1,'FaceColor',[1 .8 .4], 'LineWidth', 3),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'),ylabel('\fontsize{14}ylo = (-8/25)*(x-5)^2+10 • yhi = 11e^-^x^/^6+9'),...
title(['\fontsize{16}MTH15 • Area Between Curves',]),...
annotation('textbox',[.5 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
25Jul13','FontSize',7)
hold on
set(gca,'XTick',[xmin:2:xmax]); set(gca,'YTick',[ymin:2:ymax])
set(gca,'Layer','top')
hold off
%
subplot(1,3,1)
area(x,y2, 'FaceColor',[0 1 0], 'LineWidth', 3),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'),...
annotation('textbox',[.15 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
25Jul13','FontSize',7)
hold on
set(gca,'XTick',[xmin:2:xmax]); set(gca,'YTick',[ymin:2:ymax])
set(gca,'Layer','top')
hold off
%
xn = linspace(xmin, xmax, 500);
subplot(1,3,3)
fill([xn,fliplr(xn)],[(-8/25)*(xn-5).^2 + 10, fliplr(11*exp(-xn/6)+9)],'m'),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'),...
annotation('textbox',[.85 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
25Jul13','FontSize',7)
hold on
set(gca,'XTick',[xmin:2:xmax]); set(gca,'YTick',[ymin:2:ymax])
set(gca,'Layer','top')
hold off
%
disp('Showing SubPlot - Hit Any Key to Continue')
pause
%
clf
fill([xn,fliplr(xn)],[(-8/25)*(xn-5).^2 + 10, fliplr(11*exp(-xn/6)+9)],'m'),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'),,ylabel('\fontsize{14}ylo = (-8/25)*(x-5)^2+10 • yhi = 11e^-^x^/^6+9'),...
title(['\fontsize{16}MTH15 • Area Between Curves',]),...
annotation('textbox',[.6 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
25Jul13','FontSize',7)
hold on
set(gca,'XTick',[xmin:2:xmax]); set(gca,'YTick',[ymin:2:ymax])
set(gca,'Layer','top')
hold off
MuPAD Code
Chabot College Mathematics
19
f := 11*exp(-x/6)+9
g := (-8/25)*(x-5)^2+10
fminusg := f-g
AntiDeriv := int(fminusg, x)
ABC := int(fminusg, x=0..10)
float(ABC)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  Net Excess Profit
 The Net Excess Profit of an investment
plan over another is given by
 P ' t   P ' t dt   dP
b
a
b
1
2
a
1
dt  dP2 dt dt
• Where dP1/dt & dP2/dt are the rates of
profitability of plan-1 & plan-2
 The Net Excess Profit (NEP) gives the
total profit gained by plan-1 over plan-2
in a given time interval.
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  Net Excess Profit
 Find the net excess profit during the
period from now until plan-1 is no longer
increasing faster than plan-2:
 Plan-1 is an investment that is currently
increasing in value at $500 per day and
dP1/dt (P1’) is increasing instantaneously
by 1% per day, as compared to plan-2
which is currently increasing in value at
$100 per day and dP2/dt (P2’) is
increasing instantaneously by 2% per day
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  Net Excess Profit
 SOLUTION:
 The functions are each increasing
exponentially (instantaneously), with
dP1/dt initially 500 and
growing exponentially dP1  500e 0.01t
with k = 0.01, so that dt
 Similarly, dP2/dt is initially dP2
0.02t
 100e
100 and growing
dt
exponentially with k = 0.02, so that
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  Net Excess Profit
 ReCall the
b


NEP

dP
dt

dP
dt
dt
1
2

NEP Eqn
a
• where a and b are determined
by the time for which plan-1 is dP
dP2
1

increasing faster than plan-2,
dt t dt
that is, [a,b] includes those
times, t, such that:
 Using the Given Data
dP1
dP2

 500e0.01t  100e0.02t
dt t dt t
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
t
Example  Net Excess Profit
 Dividing Both Sides of the InEquality
0.01t
0.02t
500e
 100e
0.02t 0.01t
0.01t

5

e

e
0.01t
100e
 Taking the Natural Log of Both Side

ln 5  e
0.01t

 ln 5  0.01t
 Divide both Sides by 0.01 to Solve for t
ln 5
ln 5
t  t 
 160.94
0.01
0.01
Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  Net Excess Profit
 The plan-1 is greater than plan-2 from
day-0 to day 160.94.
 Thus after rounding the NEP covers the
time interval [0,161]. The the NEP Eqn:
161days
NEP  
0 days
500e
0.01t

 100e 0.02t dt
 Doing the Calculus
 500e
161
0
0.01t
 100e
Chabot College Mathematics
25
0.02t
161

 500 0.01t 100 0.02t 
dt  
e 
e 
0.02
 0.01
0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  Net Excess Profit
 500 0.01(161) 100 0.02(161)   500 0.01( 0) 100 0.02( 0 ) 

e

e

e

e


0.02
0.02
 0.01
  0.01
 79,999.96
 STATE: In the initial 161 days, the Profit
from plan-1 exceeded that of plan-2 by
approximately $80k
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  Net Excess Profit
 The Profit Rates
 The NEP (ABC)
MTH15 • Net Excess Profit
MTH15 • Net Excess Profit
2.5
P1'= 0.5ex/100 • P2' = 0.5e x/50 ($k)
P1'= 0.5ex/100 • P2' = 0.5e x/50 ($k)
2.5
2
1.5
dP1 dt
1
dP2 dt
0.5
0
Bruce May er, PE • 25Jul13
0
20
40
60
80
t (days)
Chabot College Mathematics
27
100
120
140
160
2
1.5
1
0.5
0
Bruce May er, PE • 25Jul13
0
20
40
60
80
100
t (days)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
120
140
160
Chabot College Mathematics
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 25Jun13
%
clear; clc; clf; % clf clears figure window
%
xmin = 0; xmax = 161;
ymin = 0; ymax = 2.5;
% The FUNCTION
x = linspace(xmin,xmax,500); y1 = .5*exp(x/100); y2 =
.1*exp(x/50);
% x in days • y's in $k
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
plot(x,y1, x,y2, 'LineWidth', 4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}t (days)'), ylabel('\fontsize{14}
P_1''= 0.5e^x^/^1^0^0 • P_2'' = 0.5e^x^/^5^0^ ($k)'),...
title(['\fontsize{16}MTH15 • Net Excess Profit',]),...
annotation('textbox',[.6 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
25Jul13','FontSize',7)
hold on
set(gca,'XTick',[xmin:20:xmax]); set(gca,'YTick',[ymin:0.5:ymax])
disp('Hit ANY KEY to show Fill')
pause
%
xn = linspace(xmin, xmax, 500);
fill([xn,fliplr(xn)],[.5*exp(xn/100), fliplr(.1*exp(x/50))],'m')
hold off
Recall: Average Value of a fcn
 Mathematically - If f is integrable on
[a, b], then the
1 b
average value of f
f ( x)dx
a
b

a
over [a, b] is

 Example  Find
3/ 2
f ( x)  x over 0,9 .
the Avg Value:
 Use Average Definition:
 
9
1 9 3/ 2
1  2x 
2 5/ 2
x
dx
 

9


0
90
9  5  0 45
5/ 2
Chabot College Mathematics
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

54

5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  GeoTech Engineering
 A Model for The rate at
dM
3

which sediment gathers at
dt
2t  3
the delta of a river is given by
• Where
– t ≡ the length of time (years) since study began
– M ≡ the Mass of sediment (tons) accumulated
 What is the average rate at which
sediment gathers during the first six
months of study?
)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  GeoTech Engineering
 By the Avg Value eqn the average rate
at which sediment gathers over the first
six months (0.5 years)
V

0.5
1 b
1
3
f t  dt  V 
dt


a
0
ba
0.5  0
2t  3
 No Integration Rule applies so try
subsitution. Let u  2t  3
d
du
du
 du
 dt
u  2t  3   2    2  dt 
dt
dt
2
 dt
 2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  GeoTech Engineering
 And u  2t  3 
u 0  20   3  0  3  3
u 0.5  20.5  3  1  3  4
 Then the Transformed Integral
 1  t 0.5 3
 
dt  V


 0.5  0  t 0 2t  3
V
 1  u  4 3 du
 


0
.
5

0

 u 3 u 2
 Working the Calculus
V  2
4
3
3 du
3 4 du
4
2 
 3ln u 3
3
2 u
2
u

V  3ln 4  ln 3  3  ln

Chabot College Mathematics
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4
 3  0.2877  0.8630

3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Example  GeoTech Engineering
 The average rate at which sediment
was gathering for the first six months
was 0.863 tons per year.
 dM/dt along
with its
average
Equal Areas
value on
[0,0.5]:
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
WhiteBoard Work
 Problems From §5.4
• P46 → Worker Productivity
• P60 → Cardiac Fluidic Mechanics
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
All Done for Today
DilBert
Integration
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.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
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
P5.4-46(b)
 Production Rates
 Cumulative Difference
• Qtot = 184/3 units
MTH15 • P5.4-46
70
Q1'= 60-2(t-1)2 • Q2' = 50-5t (units/hr)
Q1'= 60-2(t-1)2 • Q2' = 50-5t (units/hr)
MTH15 • P5.4-46
60
50
40
30
20
10
0
Bruce May er, PE • 25Jul13
0
0.5
1
1.5
2
2.5
t (Hrs after 8am)
Chabot College Mathematics
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3
3.5
4
70
60
50
40
30
20
10
0
Bruce May er, PE • 25Jul13
0
0.5
1
1.5
2
2.5
t (Hrs after 8am)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
3
3.5
4
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-25_sec_5-4_Definite_Integral_Apps.pptx