§12.5 The Fundamental Theorem of
Calculus
The student will learn about:
the definite integral,
the fundamental theorem of calculus,
and using definite integrals to find average
values.
1
§13.1 Area Between Curves
The student will learn about:
Area between a curve
and the x axis ,
area between two
curves, and
an application.
2
Negative Values
If f (x) is positive for some values of x on [a,b] and
negative for others, then the definite integral symbol
a f (x) dx
b
Represents the cumulative
sum of the signed areas
between the graph of f (x) and
the x axis where areas above
are positive and areas below
are counted negatively.
Remember area measure is
never negative!
y = f (x)
B
a
A
b
 a f (x) dx   A  B
b
3
Negative Values
The area between the graph of a negative function and
the x axis is equal to the definite integral of the
negative of the function.
A   a [  f (x) ]dx
b
B   b f (x) dx
c
y = f (x)
B
a
A
b
c
4
Area Between Curve and x axis
The area between f (x) and the x axis can be
found as follows:
For f (x)  0 over [a, b]: Area =  a f ( x) dx
b
i.e. f (x ) is above the x axis.
For f (x)  0 over [a, b]: Area =  a [  f (x) ]dx
b
i.e. f (x ) is below the x axis.
5
Example
Find the area bounded by y = 4 – x 2; y = 0, 1  x  4.
Area = A 1 - A 2
A1
  0 f (x) dx   2 f (x) dx
2
4
A2
x 3 2 
x 3  4
 4x 
| 0   4x 
|
 2
3
3


= 8 – 8/3 – (16 – 64/3 – 8 + 8/3) = 16
6
Area Between Two Curves
Theorem 1. If f (x)  g (x) over the interval
[a, b], then the area bounded by y = f (x) and
y = g (x) for
a  x  b is given by:
A   a  f (x)  g (x) dx
b
f (x)
A
a
b g (x)
7
Example
Find the area bounded by y = x 2 - 1; y = 3.
Note the two curves intersect at
–2 and 2 and y = 3 is the larger
function on –2  x  2.
A    2 3  (x2  1) dx
2
= 3x –
x3
2
/3 + x |  2 = 6 – 8/3 + 2 – (-6 + 8/3 – 2)
32

 10.67
3
8
Example
Find the area bounded by y = x 2 - 1; y = 3.
Note many graphing
calculators have numeric
integration routines.
A    2 3  (x2  1) dx
2
   2 [4  x2 ]dx
2
= 10.667
9
Income Distribution - Application
The U.S. Bureau of the Census compiles data on
distribution of income among families. This data can be
fitted to a curve using regression analysis. This curve is
called a Lorenz curve. The variable x represents the
cumulative percentage of families at or below the given
income level. The variable y represents the cumulative
percentage of total family income received. Absolute
equality of income would occur if the area between the
Lorenz curve and y = x were 0. The maximum possible
area between the Lorenz curve and y = x is ½ the area of
the triangle below y = x indicating absolute inequality of
income (i.e. one family has all the income).
10
Income Distribution - Application
The ratio of the area bounded by y = x and the Lorenz
curve is the area of the triangle under the line y = x form
x = 0 to x = 1 is called the index of income concentration.
Gini index of Income Concentration
If y = f (x) is the Lorenz curve, then
Index of income concentration = 2  0 x  f (x) dx
1
A measure of 0 indicates absolute equality. A measure
of 1 indicates absolute inequality.
11
Example
A country is planning changes in tax structure in
order to provide a more equitable distribution of
income. The two Lorenz curves are:
f (x) = x 2.3 currently and g (x) = 0.4x + 0.6x 2 proposed.
Will the proposed changes work?
Currently: Gini index of income concentration =
2
3. 3
x
x
1

 1
2.3
 0.3939
2  0 [ x  x ]dx  2  

3.3  0
 2
|
Future: Gini index of income concentration =
2
3
1
0
.
6
x
0
.
6
x
2


2  0 [ x  (0.4 x  0.6 x ]dx  2 

  0.20
3  0
 2
12
1
|
Summary.
We reviewed the definite integral as the area
between a curve and the x axis.
We learned how to calculate area when the
curve was below the x axis.
We learned how to calculate the area between
two curves.
We learned about the Lorenz curve and the
Gini index of income concentration.
13
Practice Problems
§7.1; 1, 3, 7, 11, 15,
17, 21, 27, 31, 35, 39,
43, 47, 51, 55, 59, 61,
65, 69, 75, 77.
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§13.2 Applications
The student will learn about:
Probability density
functions , Continuous
income stream,
Future value of a
continuous income
stream, and
Consumers’ and
producers’ surplus.
15
Probability Density Function
A probability density function must satisfy:
1. f (x)  0 for all x
2. The area under the graph of f (x) = 1
3. If [c, d] is a subinterval then
Probability (c  x  d) = c f (x) dx
d
Probability density function
16
Example
In a certain city the daily use of water in hundreds of
gallons per household is a continuous random variable
with probability density function
f (x)  .15 e
 .15x
if x  0. Otherwise f (x)  0
Find the probability that a household chosen at random
will use between 300 and 600 gallons.
Probability (3  x  6) =

6
3
.15 e
= e
 .15 x
 0.15 x
dx
6
|3
= - e – . 9 + e - . 45  .23
17
Continuous Income Stream
Total Income for a Continuous Income Stream.
If f (t) is the rate of flow of a continuous income stream,
the total income produced during the time period from
t = a to t = b is:
Total income =
 a f (t ) dt
b
a
Total Income
b
18
Example
Find the total income produced by a continuous
income stream in the first 2 years if the rate of
flow is f (t) = 600 e 0.06 t
Total income =  0 600 e
2
0.06t
dt
= 10,000 e 0.06 t
2
|0
= 10,000 (e 0.12 – 1)
 $1,275
19
Future Value
of a Continuous Income Stream
From the previous example and previous work we are
familiar with the continuous compound interest formula
A = Pert.
Future Value of a Continuous Income Stream
If f (t) is the rate of flow of a continuous income stream,
0  t  T, and if the income is continuously invested at a
rate r compounded continuously, the the future value,
FV, at the end of T years is given by
FV   0 f (t ) e
T
r (T  t)
dt  e  0 f (t ) e dt
rT
T
r t
20
Example
Let’s continue the previous example where
f (t) = 600 e 0.06 t
Find the Future Value in 2 years at a rate of 10%.
FV  e  0 f (t ) er t dt
T
rT
e
0.10( 2 )

 600 e
2
0
0.2
600 e0.06t e  0.10t dt

2
0
r = 0.10, T = 2, f (t) = 600 e 0.06t
e 0.04t dt
= (600) (1.22140) (1.92209) = 1,408.59
21
22
Consumers’ Surplus
If (x, p) is a point on the graph of the price-demand
equation p = D (x), then the consumers’ surplus, CS, at a
x
price level of p is
CS   D (x)  p dx
0
Which is the area between p = p and p = D (x) from
x = 0 to x = x. The consumers’ surplus represents the
total savings to consumers who are willing to pay more
than p for the product but are still able to buy the
product for p.
CS
p
x
23
Example
Find the consumers’ surplus at a price level of p = $120
for the price-demand equation
p = D (x) = 200 – 0.02x
Step 1. Find x, the demand when the price is p = 120:
p = 200 – 0.02x
120 = 200 – 0.02x
x = 4000
24
Example - continued
Find the consumers’ surplus at a price level of p = $120
for the price-demand equation
x = 4000
p = D (x) = 200 – 0.02x
Step 2. Find the consumers’ surplus:
CS   0 D (x)  p dx
x
 0
4000
 0
(200  0.02x  120) dx
4000
(80  0.02x) dx
= 80x –
0.01x 2
4000
|0
= 320,000 – 160,000 = $160,000
25
Producers’’ Surplus
If (x, p) is a point on the graph of the price-supply
equation p = S (x), then the producers’ surplus, PS, at a
price level of p is PS  x p  S (x) dx
0
which is the area between p = p and p = S (x) from x = 0 to
x = x. The producers’ surplus represents the total gain to
producers who are willing to supply units at a lower price
than p but are still able to supply units at p.
CS
p
x
26
Example
Find the producers’ surplus at a price level of p = $55
for the price-supply equation
p = S (x) = 15 + 0.1 x + 0.003 x 2
Step 1. Find x, the supply when the price is p = 55:
p = 15 + 0.1 x + 0.003 x 2
55 = 15 + 0.1 x + 0.003 x 2
0 = 0.003 x 2 +0.1 x - 40
Solving for x using a graphing utility.
x = 100
27
Example - continued
Find the producers’ surplus at a price level of p = $55
for the price-supply equation
x = 100
p = S (x) = 15 + 0.1 x + 0.003 x 2
Step 2. Find the producers’ surplus:
PS   0 p  S (x) dx
100
2
  0 [55  (15  0.1 x  0.003 x ) ] dx
x
  0 (40  0.1 x  0.003 x2 ) dx
100
= 40x –
0.05x 2
– 0.001
x3
100
|0
= 4,000 – 500 – 1,000 = $2,500
28
Summary.
We learned how to use a probability density function.
We defined and used a continuous income stream.
We found the future value of a continuous income
stream.
We defined and calculated a consumer’s surplus.
We defined and calculated a producer’s surplus.
29
Practice Problems
§7.2; 1, 3, 5, 7, 9, 13,
17, 21, 29, 33, 37, 41,
43, 47, 51.
30