# Theorem of Calc §5.3 Fundamental Chabot Mathematics Bruce Mayer, PE

```Chabot Mathematics
&sect;5.3 Fundamental
Theorem of Calc
Bruce Mayer, PE
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Review &sect;
5.2
• &sect;5.2 → AntiDerivatives by Substitution
• &sect;5.22 → HW-23
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
&sect;5.3 Learning Goals
 Show how area under a curve can be
expressed as the limit of a sum
 Deﬁne the deﬁnite integral and explore its
properties
 State the fundamental theorem of calculus,
and use it to compute deﬁnite integrals
 Use the fundamental theorem to solve applied
problems involving net change
 Provide a geometric justiﬁcation of the
fundamental theorem
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Area Under the Curve (AUC)
 The AUC has many Applications in Business,
Science, and Engineering
Calculation of
Geographic Areas
Chabot College Mathematics
4
River Channel
Cross Section
Wind-Force
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Area Under Function Graph
 For a Continuous Function, approximate
the area between the Curve and the
x-Axis by Summing Vertical Strips
• Use Rectangles of Equal Width
– Three Possible Forms
Left end points
Right end points
y  f ( x)
y  f ( x)
Midpoints
y  f ( x)
Strip Width
x 
x
ba
n
(n strips)
Chabot College Mathematics
5
a
b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example: Strip Sum
 Approximate the area
2
f
(
x
)

2
x
on
0,
2


under the graph of
MTH15 • Area by Strip Addition
8
7
 Use
• Strip
MidPoints
6
y = f(x) = 2x2
• n=4
(4 strips)
Bruce May er, PE • 24JUul13
5
4
3
2
1
0
Chabot College Mathematics
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0
0.5
1
x
1.5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
2
Example: Strip Sum GamePlan
MTH15 • Area by Strip Addition
8
Bruce May er
er, •PE
24Jul13
• 24JUul13
7
y = f(x) = 2x2
6
5
4
3
2
1
0
0
Chabot College Mathematics
7
0.5
1
x
1.5
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 24Jul13
% XY_Area_fcn_Graph_6x6_BlueGreen_BkGnd_Template_1306.m
%
% The FUNCTION
xmin = 0; xmax = 2;
ymin = 0; ymax = 8;
% The FUNCTION
x = linspace(xmin,xmax,20); y = 2*x.^2;
%
% 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 .8 1] , 'LineWidth', 3), axis([xmin xmax
ymin ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = f(x)
= 2x^2'),...
title(['\fontsize{16}MTH15 • Area by Strip Addition',]),...
annotation('textbox',[.13 .82 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
24JUul13','FontSize',7)
hold on
set(gca,'Layer','top')
plot(x,y, 'LineWidth', 3),
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 24Jul13
%
% The Limits
xmin = 0; xmax = 2;
ymin = 0; ymax = 8;
% The FUNCTION
x = linspace(xmin,xmax,500); y = 2*x.^2;
x1 = [0.25:.5:1.75]; y1 = 2*x1.^2
%
% 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
plot(x,y, 'LineWidth', 4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = f(x) = 2x^2'),...
title(['\fontsize{16}MTH15 • Area by Strip Addition',]),...
annotation('textbox',[.13 .82 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'Bruce
Mayer • 24Jul13','FontSize',7)
hold on
area([(x1(1).25)*ones(1,100),(x1(1)+.25)*ones(1,100)],[y1(1)*ones(1,100),y1(1)*ones(1,100)],'FaceColor',[1 .8 1])
area([(x1(2).25)*ones(1,100),(x1(2)+.25)*ones(1,100)],[y1(2)*ones(1,100),y1(2)*ones(1,100)],'FaceColor',[1 .8 1])
area([(x1(3).25)*ones(1,100),(x1(3)+.25)*ones(1,100)],[y1(3)*ones(1,100),y1(3)*ones(1,100)],'FaceColor',[1 .8 1])
area([(x1(4).25)*ones(1,100),(x1(4)+.25)*ones(1,100)],[y1(4)*ones(1,100),y1(4)*ones(1,100)],'FaceColor',[1 .8 1])
plot(x,y, 'LineWidth', 4)
set(gca,'Layer','top')
plot(x1,y1,'g d', 'LineWidth', 4)
plot([x1(1)-.25,x1(1)+.25],[y1(1),y1(1)], 'm', [x1(2)-.25,x1(2)+.25],[y1(2),y1(2)], 'm',...
[x1(3)-.25,x1(3)+.25],[y1(3),y1(3)], 'm', [x1(4)-.25,x1(4)+.25],[y1(4),y1(4)], 'm','LineWidth',2)
plot([x1(1)-.25,x1(1)-.25],[0,y1(1)], 'm',[x1(2)-.25,x1(2)-.25],[0,y1(2)], 'm',...
[x1(3)-.25,x1(3)-.25],[0,y1(3)], 'm', [x1(4)-.25,x1(4)-.25],[0,y1(4)], 'm', 'LineWidth', 2)
set(gca,'XTick',[xmin:0.5:xmax]); set(gca,'YTick',[ymin:1:ymax])
MTH15 • Area by Strip Addition
8
Example: Strip Sum
7
6
y = f(x) = 2x2
 The Algebra
Bruce May er
er, •PE
24Jul13
• 24JUul13
5
4
3
2
1
midpoints
0
0
0.5
1
x
A  x  f (m1 )  f (m2 )  f (m3 )  f (m4 )
1
A f
2
1
 
 4
 3
f  
 4
5
f  
 4
 7 
f  
 4 
1  1 9 25 49  21
A     
2 8 8 8
8 4
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
1.5
2
Area under a Curve
y  f ( x)
ba
Width: x 
n
(n strips)
x
a
b
 GOAL: find the exact area under the
graph of a function; i.e., the curve
 PLAN: Use an infinite number of strips
of equal width and compute their area
with a limit.
Chabot College Mathematics
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Area Under a Curve
f xk 
 Function, f(x), on
interval [a,b] is:
y  f ( x)
• Continuous
• NonNegative
a
xk
 Then the Area Under the Curve, A
b
A  lim  f ( x1 )  f ( x2 )  ...  f ( xn ) x
n
 The x1, x2, …, xn-1,xn are arbitrary, n
SubIntervals each with width (b − a)/n
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Riemann Sum ∑f(xk)&middot;∆x
 For a Continuous, NonNeg fcn over
[a,b] divided into n-intervals of Equal
Width, ∆x = (b−a)/n, The AUC can be
approximated by the sum the area of
Vertical Strips
AUC  A1  A2    Ak  An1  An
AUC  f x1  x  f x2  x    f xk  x  f xn1  x  f xn  x
n
AUC   HEIGHT k  ConstantWi dth 
AUC 
  f x  x
k 1
Chabot College Mathematics
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k 1
n
k

n
  f x  x
k 1
k
Riemann ∑
for
x  Const
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Riemann ∑ → Definite Integral
 For a Continuous, NonNeg fcn over [a,b]
divided into n-intervals of Equal Width, ∆x
= (b−a)/n, The AUC can be calculated
EXACTLY by the Riemann sum as the
number of strips becomes infinite.
 This Process of finding an Infinite Sum is
called “Integration”;
• &quot;to render (something) whole,&quot; from Latin
integratus, past participle of integrare &quot;make
whole,&quot;
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Riemann ∑ → Definite Integral
 As the No. of Strips increase the AUC
Calculation becomes more accurate
MTH15 • Area by Strip Addition
MTH15 • Area by Strip Addition
8
Bruce May er • 24Jul13
7
y = f(x) = 2x2
4
3
5
4
3
2
2
1
1
0
0.5
1
Fifty Strips
6
5
0
Bruce May er • 24Jul13
7
Twenty Strips
6
y = f(x) = 2x2
8
1.5
0
2
0
0.5
1
x
x
1.5
 The
Riemann-Sum
to
Definite-Integral
n
n
b

b  a
lim   f  xk 
 lim   f xk  x   f  x   dx
a
n 
n 
k 1
Chabot College Mathematics
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n
k 1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
2
% 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
plot(x,y, 'LineWidth', 4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = f(x)
= 2x^2'),...
title(['\fontsize{16}MTH15 • Area by Strip Addition',]),...
annotation('textbox',[.13 .82 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'Bruce Mayer •
24Jul13','FontSize',7)
hold on
bar(x1,y1, 'BarWidth',1, 'FaceColor', [1 .8 1], 'EdgeColor','b',
'LineWidth', 2)
set(gca,'XTick',[xmin:0.5:xmax]); set(gca,'YTick',[ymin:1:ymax])
set(gca,'Layer','top')
plot(x,y, 'LineWidth', 3)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 24Jul13
%
% The Limits
xmin = 0; xmax = 2;
ymin = 0; ymax = 8;
% The FUNCTION
x = linspace(xmin,xmax,500); y = 2*x.^2;
x1 = [1/20:1/10:39/20]; y1 = 2*x1.^2;
Definite Integral Symbology
upper limit of integration
Integration
Symbol

f  x  dx
b
a
integrand
lower limit of integration
variable of integration
(dummy variable)
It is called a dummy variable because
the answer does not depend on the
symbol chosen; it depends only on a&amp;b
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx

Recall Fundamental Theorem
 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
18
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-24_sec_5-3_Fundamental_Theorem.pptx
Fundamental Theorem – Part2
 Previously we stated that the
AntiDerivative of f(x) is F(x), so then
d
d
F x  
dx
dx
 f xdx  d  f x  d   f x  1 f x  f x
 Now consider the definite Integral
(AUC) Relationship to the AntiDerivative
 f x dx
b
a
Chabot College Mathematics
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 F b  F a  
F xa
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
b
DefiniteIntegral↔AntiDerivative
 f x dx
b
a
 F b  F a  
F xa
b
 That is, The AUC for a continuous
Function, f(x), spanning domain [a,b] is
the AntiDerivative evaluated at b minus
the AntiDerivative evaluated at a.
– D. F. Riddle, Calculus and
Analytical Geometry, Belmont,
pp. 179-181, pg. 770
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example  Find AUC
 Find the area under the graph of y = 2x3
2
0
Gives the area since 2x3
is nonnegative on [0, 2].
2x 3dx
 Then
2
0
2
Antiderivative
Chabot College Mathematics
21
     8 sq.units
1 4
1 4 1 4
2 x dx  x
 2  0
2 0
2
2
3
Fund. Thm.
of Calculus
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Rules for Definite Integrals
1. Constant Rule:
for any constant, k
k
dx

k

x

C

b
b
b
2. Sum/Diff
4.   f ( x)  g ( x)  dx   f ( x)dx   g ( x)dx
a
a
a
Rule:
 f x dx  0
a
3. Zero Width
Rule
4. Domain
Reversal
2.
Rule
Chabot College Mathematics
22

a
b
a
a
f ( x)dx   f ( x)dx
b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
.
Rules for Definite Integrals
5. SubDivision Rule, for (a&lt;b&lt;c)

b
a
c
b
a
c
f ( x)dx   f ( x)dx   f ( x)dx
y
a
Chabot College Mathematics
23
b
c
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example  Eval Definite Integral
 Find a Value for

5
1
1 

 2 x   1 dx
x 

 The Reduction using the Term-by-Term
rule
5
5
1 
2
1  2 x  x  1dx  x  ln x  x1

 

 52  ln 5  5  12  ln1  1
 28  ln5  26.39056
Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example  Def Int by Substitution

Find:
1
x 1
1 2 Clarify
12
2
2
 2 x  3 x  3x dx
Limits 
 2x  3 x  3x 

0


x 0

let u  x 2  3 x
 Let:
 Then find dx(du) and u(x=0), and u(x=1)




d
d
d 2
2
u  x  3x 
u  x  3x  u 
x  3x  2 x  3
dx
dx
dx
2
x u  x  3x
du
 du 2 x  3  dx
2


dx

 dx
2 x  3 0 0  3  0  0
1  2 x  3
1 12  3 1  4
2
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example  Def Int by Substitution
 SubOut x2+3x, and the Limits
 2 x  3x
x 1
2
x 0
 3x

12
u 4

2 x  3u 
u 0
dx  
12
du
2 x  3
3 2 u 4
 Dividing
u 4
u 
2 3 2 u 4
12
out the  u 0 u  du   3 2   3 u u 0

 u 0
2x+3
2 32
2
2
2 8 16
32
3
 Then 4  0  
4 
64   
3 
3
 Thus Ans
Chabot College Mathematics
26
 2 x  3x
1
0
2
 3x
2

12
3 1
16

3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
3
The Average Value of a Function
MTH15 • Meaning of Avg
350
300
250
y = f(x)
 At y = yavg
there at
EQUAL
AREAS
above &amp;
below the
Avg-Line
Avg Line
200
150
100
50
0
Chabot College Mathematics
27
Bruce May er, PE • 24Jul13
0
2
4
6
8
x
10
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
14
16
Chabot College Mathematics
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 24Jul13
% Area_Between_fcn_Graph_BlueGreen_BkGnd_Template_1306.m
% Ref: E. B. Magrab, S. Azarm, B. Balachandran, J. H. Duncan, K. E.
% Herhold, G. C. Gregory, &quot;An Engineer's Guide to MATLAB&quot;, ISBN
% 978-0-13-199110-1, Pearson Higher Ed, 2011, pp294-295
%
clc; clear
% The Function
xmin = 0; xmax = 16;
ymin = 0; ymax = 350;
xct = 1000
x = linspace(xmin,xmax,xct);
y1 = .5*x.^3-9*x.^2+11*x+330;
yavg = mean(y1)
y2 = yavg*ones(1,xct)
%
%
% Find Zeros
plot(x,y1, x,y2, 'k','LineWidth', 2), axis([0 xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = f(x)'),...
title(['\fontsize{16}MTH15 • Meaning of Avg',]),...
annotation('textbox',[.13 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor',
'none', 'String', 'Bruce Mayer, PE • 24Jul13','FontSize',7)
display('Showing 2Fcn Plot; hit ANY KEY to Continue')
% &quot;hold&quot; = Retain current graph when adding new graphs
hold on
%
nct = 500
xn = linspace(xmin, xmax, nct);
fill([xn,fliplr(xn)],[.5*xn.^3-9*xn.^2+11*xn+330,
fliplr(yavg*ones(1,nct))],'m'),grid
plot(x,y1), grid
The Average Value of a Function
 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-24_sec_5-3_Fundamental_Theorem.pptx
Net Change
 If the Rate of Change (RoC), dQ/dx =
Q’(x) is continuous over the interval
[a,b], then the NET CHANGE in Q(x) is
Given by
Qb  Qa    dQ dxdx   Q' x dx
Chabot College Mathematics
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b
b
a
a
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example  Find Net Change
 A small importer of Gladiator
merchandise has modeled her monthly
profits since the company was created
on January 1, 1997 by the formula
P(t) = 0.7t - 7.7t + 26.6t - 28t
4
3
2
• Where
– P ≡ \$-Profit in 100’s of Dollars (\$c or c-Notes)
– t ≡ year of operation for the company
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example  Find Net Change
 What is the importer’s net change in
profit between the beginning of the
years 2000 and 2003?
 SOLUTION:
 Recall t is in years after 1997, Thus
• Year 2000 corresponds to t = 3
• Year 2003 corresponds to t = 6
 Then in this case the Net Change in
Profit over [3,6] →
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Example  Find Net Change

6
3

 0.14t

Pt    0.7t 4  7.7t 3  26.6t 2  28t  dt
6
3
5
 1.925t  8.667t  14t
4
3

2 6
3
= &eacute;&euml;0.14(6)5 -1.925(6)4 + 8.667(6)3 -14(6)2 &ugrave;&ucirc;
-&eacute;&euml;0.14(3)5 -1.925(3)4 + 8.667(3)3 -14(3)2 &ugrave;&ucirc;
 \$13.545c
 Thus Her monthly profits increased by
about \$1,354.50 between 2000 &amp; 2003
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
WhiteBoard Work
 Problems From &sect;5.3
• P74 → Water Consumption
• P80 → Distance Traveled
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
All Done for Today
Students
Should Calc
52 x 7 2  66 x 5 2  22 x 3 2
dx
0
x
1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
All Done for Today
Fundamental
Theorem
Part-1
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Fundamental Theorem Proof
Let
Aa  x   area under the
curve from a to x.
(“a” is a constant)
a
x xh
Aa  x  Ax  x  h 
Aa  x  h 
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Then:
Aa  x   Ax  x  h   Aa  x  h 
Ax  x  h   Aa  x  h   Aa  x 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
min f
max f
xh
x
h
The area of a rectangle drawn
under the curve would be less
than the actual area under the
curve.
The area of a rectangle drawn
above the curve would be
more than the actual area
under the curve.
short rectangle  area under curve  tall rectangle
h  min f  Aa  x  h   Aa  x   h  max f
min f 
Chabot College Mathematics
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Aa  x  h   Aa  x 
h
 max f
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx

min f 
Aa  x  h   Aa  x 
h
 max f
As h gets smaller, min f and max f get closer together.
lim
h 0
Aa  x  h   Aa  x 
h
 f  x
0  F a  c
initial
value
Take the anti-derivative
of both
sides to find an explicit formula
for area.
40
of derivative!
Aa  a   F  a   c
d
Aa  x   f  x 
dx
Chabot College Mathematics
Aa This
x  isFthe xdefinition
c
F  a   c
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx

min f 
Aa  x  h   Aa  x 
h
 max f
As h gets smaller, min f and max f get closer together.
lim
h 0
Aa  x  h   Aa  x 
h
 f  x
Aa  x   F  x   c
Aa  a   F  a   c
d
Aa  x   f  x 
dx
0  F a  c
Aa  x   F  x   F  a 
F  a   c
Area under curve from a to x = antiderivative at x minus
Bruce Mayer, PE
Chabot College Mathematics
antiderivative at a.
41
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.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-24_sec_5-3_Fundamental_Theorem.pptx
x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-24_sec_5-3_Fundamental_Theorem.pptx
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