Chabot Mathematics
§5.1
Integration
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Review §
4.4
 Any QUESTIONS About
• §4.4 → Exp & Log
Math Models
 Any QUESTIONS
About HomeWork
• §4.4 → HW-21
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
§5.1 Learning Goals
 Define AntiDerivative
 Study and compute
indefinite integrals
 Explore differential
equations and
Initial/Boundary
value problems
 Set up and solve VariableSeparable differential equations
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Fundamental Theorem of Calculus
 The fundamental theorem* of calculus
is a theorem that links the concept of the
derivative of a function with the concept
of the integral.
• Part-1: Definite Integral
(Area Under Curve)
 f xdx  F b  F a
b
a
• Part-2: AntiDerivative
if
F x    f  x dx then
Chabot College Mathematics
4
d
F x   d
dx
dx
 f xdx  f x
* The Proof is Beyond the Scope of MTH15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
AntiDifferentiation
 Using the 2nd Part dF
 f  x  or F  x    f x dx
of the Theorem
dx
 F(x) is called the AntiDerivative of f(x)
• Example:
Find f(x)
when
df  x 
 4x 3
dx
• ONE Answer is f x   x
4
 
d
d 4
3
• As Verified by
f x  
x  4x
dx
dx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Fundamental Property of Antiderivs
 The Process of Finding an
AntiDerivavite is Called: InDefinite
Integration
 The Fundamental Property of
AntiDerivatives:
• If F(x) is an AntiDerivative of the
continuous fcn f(x), then any other
AntiDerivative of f(x) has the form
G(x) = F(x) + C, for some constant C
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Fundamental Property of Antiderivs
 Proof of G(x) = F(x) + C
 Assertion: both G(x) & F(x)+C are
AntiDerivatives of f(x); that is:
d
d
G x   f x   F x   C 
dx
dx
? d
d
 Using
G x   F x   C  Transitive Property
dx
Derivative dx dG ? dF
dC


Derivative of a Sum
Rules
dx dx dx
dG ? dF
Derivative of a Const

0
dx dx
dG dF
f x  

 f x 
dx dx
Bruce Mayer, PE
Chabot College Mathematics

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BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
The Indefinite Integral
 The family of ALL AntiDerivatives of f(x)
is written
 f ( x) dx  F ( x)  C
 The result of ∫f(x)dx is called the
indefinite integral of f(x)
3
4


u
x
dx

4
x
dx

x
C
 Quick Example for: 

• u(x) has in INFINITE
NUMBER of Results,
Two Possibilities:
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F x   x 4
or
G x   x 4  2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
The Meaning of “C”
 The Constant, C, is the y-axis “Anchor
Point” for the “natural Response” fcn
F(x) for which C = 0.
• C is then the y-intercept
of F(x)+C; i.e.,
G0  F 0  C
 Adding C to F(x) creates a “family” of
functions, or curves on the graph, with
the SAME SHAPE, but Shifted
VERTICALLY on the y-axis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
The Meaning of “C” Graphically
MTH15 • Familiy of AntiDerivatives
y = G(x) = F(x)+C = 7e-5x/2 + 5x - 8 + C
20
15
10
5
0
-5
-10
-4
Chabot College Mathematics
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B. May er • 20Jul13
-3
-2
-1
0
x
1
2
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
4
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 20Jul13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
% The Limits
xmin = -4; xmax = 4;
ymin = -10; ymax = 20;
% The FUNCTION
x = linspace(xmin,xmax,1000); y = 7*exp(-x/2.5) + 5*x -8;
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg(['white']) % whitebg([0.8 1 1]); % Chg Plot BackGround to
Blue-Green
plot(x,y, x,y+9,x,y-pi,x,y+sqrt(13),x,y-7, 'LineWidth',
4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = G(x) =
F(x)+C = 7e^-^5^x^/^2 + 5x - 8 + C'),...
title(['\fontsize{16}MTH15 • Familiy of AntiDerivatives',]),...
annotation('textbox',[.71 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'B. Mayer • 20Jul13','FontSize',7)
hold on
plot(zxv,zyv, 'k', zxh,zyh, 'k', [-1.4995, -1.4995], [ymin,ymax],
'--m', 'LineWidth', 2)
set(gca,'XTick',[xmin:1:xmax]); set(gca,'YTick',[ymin:5:ymax])
MuPAD Code
Bruce Mayer, PE
MTH15 20Jul13
F(x) = 7*exp(-2*x/5) + 5*x -8
f(x) = int(G, x)
G := 7*exp(-2*x/5) + 5*x -8
dgdx := diff(G, x)
assume(x > -6):
xmin := solve(dgdx, x)
xminNo := float(xmin)
Gmin := subs(G, x = xmin)
GminNo := float(Gmin)
plot(G, x=-4..4, GridVisible = TRUE,
LineWidth = 0.04*unit::inch)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Evaluating C by Initial/Boundary
 A number can be found for C if the
situation provides a value for a SINGLE
known point for G(x) →
(x, G(x)); e.g., (xn, G(xn)) = (73.2, 4.58)
• For Temporal (Time-Based) problems the
known point is called the INITIAL Value
– Called Initial Value Problems
• For Spatial (Distance-Based) problems the
known point is called the BOUDARY Value
– Called Boundary Value Problems
Chabot College –
Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Common Fcn Integration Rules
1. Constant Rule:
for any constant, k
k
dx

k

x

C

n 1
x
 x dx  n  1  C
1
3. Logarithmic Rule:
dx

ln
x

C
x
for any x ≠ 0
1 kx
4. Exponential Rule:
kx
e
dx

e

C

for any constant, k
k
2. Power Rule:
for any n ≠ −1
Chabot College Mathematics
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n
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Integration Algebra Rules
1. Constant Multiple Rule: For any
constant, a




a

u
x
dx

a

u
x
dx


2. The Sum or Difference Rule:










u
x

v
x
dx

u
x
dx

v
x


 dx










p
x

q
x
dx

p
x
dx

q
x
dx



•
This often called the Term-by-Term Rule
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Use the Rules
 Find the family of
AntiDerivatives
corresponding to
x
2
 2 x  1 dx
 SOLUTION:
 First Term-by-Term → break up each
term over addition and subtraction:
x

2
x

1
dx

x
dx

2
x
dx

1
dx




2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Use the Rules
 Move out the constant in the 2nd integral
(2), and state sqrt as fractional power
x
2
 2 x  1 dx   x dx  2 x
2
12
dx   1 dx
 Using the Power Rule
2 1
1/ 2 1
0 1
x
x
x
2
1/ 2
x
dx

2
x

 dx  1 dx  2  1  2  1 / 2  1  0  1  C
 Cleaning 2
x3 4 3/ 2
 x  2 x  1 dx  3  3  x  x  C
Up →
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Propensity to Consume
 The propensity to consume (PC) is the
fraction of income dedicated to
spending (as opposed to saving).
 A Math Model for the marginal
propensity to consume (MPC) for a
certain population:
• Where
MPC x   e
0.8 x
– MPC is the rate of change in PC
– x is the fraction of income that is disposable.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Propensity to Consume
 If the propensity to consume is 0.8
when disposable income is 0.92 of total
income, find a formula for PC(x)
 SOLUTION:
 From the Problem Statement that the
MPC is a marginal function discern that
d
MPC  x  
PC  x ,
dx
 Thus the PC fcn is the AntiDerivative of
MPC(x)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Propensity to Consume
 Find PC by
Integrating
PC x    MPC x  dx
𝑀𝑃𝐶 =
𝑒−0.8𝑥
e 0.8 x dx
1 0.8 x

e
C
 0 .8
 1.25e 0.8 x  C
 This is satisfactory for a general
solution, but need the particular
solution so that PC(0.92) = 0.8
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Propensity to Consume
 Use the (x,PC) = (0.92,0.8) Boundary
Value to Find a NUMBER for the
Constant of Integration, C
PC(x) = -1.25e-0.8x + C
0.8 = -1.25e-0.8(0.92) +C
C = 0.8+1.25e-0.8(0.92)  1.4
 With C ≈ 1.4, state the particular
solution to this Boundary Value Problem
PCx   1.25e
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0.8 x
 1.4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Differential Equations (DE’s)
 A Differential Equation is an equation
that involves differentials or derivatives,
and a function that satisfies such an
equation is called a solution
 A Simple Differential
dy
Equation is an
equation which includes dx
two differentials in the form
of a derivative
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 f (x )
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Differential Equations (DE’s)
 For some function f. Such a Simple
Differential Equation can be solved by
integrating:
dy
 dy

 f x     f x  dx 
dx
 dx

dy dx
  f x   dx
dx 1
dy  f x   dx   dy   f x   dx   1 dy   f x   dx  y   f ( x) dx
 In summary the Solution, y, to a Simple
DE can be found
y

f
(
x
)
dx
by the integration

Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Simple DE
d
 From the
PC x   e 0.8 x
Previous Example dx
 As previously solved for the general
solution
0.8 x
PC x   1.25e
C
by Integration:
 Then used the Boundary Value,
(0.92, 0.8), to find the Particular Solution
PCx   1.25e 0.8 x  1.4
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Variable-Separable DE’s
 A Variable Separable Differential
equation is a differential dy f (x)
=
equation of the form
dx g(y)
• For some integrable functions f and g
 Such a differential equation can be
solved by separating the single-variable
functions and integrating:
g  y   dy f x 

 g  y  dy  f x dx   g  y  dy   f x  dx


dx  dx g  y  
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Fluid Dynamics
 The rate of change in volume (in cubic
centimeters) of water in a draining
container is proportional to the square
root of the depth (in cm) of the water
after t seconds, with constant of
proportionality 0.044.
 Find a model for the volume of water
after t seconds, given that initially the
container holds 400 cubic centimeters.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Fluid Dynamics
 SOLUTION:
 First, TRANSLATE the written
description into an equation:
• “rate of change d

V t 
in volume”
dt
• “is proportional to the
k
V
square root of volume”
• “with constant of
proportionality equal to 0.044” k  0.044
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Fluid Dynamics
 So the (Differential) dV
 0.044 V
Equation
dt
 Note that the right side does not
explicitly depend on t, so we can’t
simply integrate with respect to t.
• Instead move the expression dV
= 0.044 dt
containing V to the left side:
V
 The Variables are now Separated,
allowing simple integration
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Fluid Dynamics
 Integrating
ò
òV
-1/2
dV
= ò 0.044 dt
V
dV = ò 0.044 dt
2V 1/2 + C1 = 0.044t + C2
V 1/2 = 0.022t + C
 Where
1
C  C2  C1 
2
 Squaring
2
Both Sides V  0.022t  C 
Find:
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
Example  Fluid Dynamics
 For The particular solution find the a
number for C using the Initial Value:
when t = 0, V = 400 cc:
• Sub (0,400) into 400 = ( 0.022(0) + C ) 2
DE Solution
C   400  20
 Thus the volume of water in the
Draining Container as a fcn of time:
V t   0.022t  20 
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
WhiteBoard Work
 Problems From §5.1
• P58 → Oil Production
(not a Gusher…)
• P73 → Car Stopping
Distance
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
All Done for Today
LOTS more
on DE’s
in MTH25
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.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-22_sec_5-1_Integration.pptx
ConCavity Sign Chart
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
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−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-22_sec_5-1_Integration.pptx