Class 9:
Area, Consumer Surplus, Integration
Demand Function
D(q)
Revenue
D(q)
q
-1.2
-10
q
What is Total Possible Revenue?
Demand Function
Total Possible
Revenue
-1.2
-8
The total
possible revenue is the money that the
producer would receive if everyone who wanted
the good, bought it at the maximum price that
he or she was willing to pay.
Consumer Surplus
Consumer
Surplus
Demand Function
D(q)
Revenue
Not
Sold
-1.2
q
-8
The total extra amount of money that people
who bought the good would have been willing
to pay is called the consumer surplus.
Finding Areas
What is the area of the region R that is
enclosed between the x-axis and the graph of
f(x) = 2x  x2/2, for x between 1 and 4?
2
1
R
1
4
Finding Areas
What is the area of the region R that is
enclosed between the x-axis and the graph of
f(x) = 2x  x2/2, for x between 1 and 4?
2
1
R
1
4
For n = 6 rectangles
2
2
f (m1)
1
1
1.5 2 2.5 3 3.5 4
{
1
4
1 m1 m2 m3 m4 m5 m6
x0 x1 x2 x3 x4 x5 x6
Dx
Sum of areas called S6
= f(m1)Dx + f(m2)Dx + f(m3)Dx + f(m4)Dx + f(m5)Dx + f(m6)Dx
= 4.531250.
More rectangles: Larger n
2
n
1
Sn 
 f ( m i )  Dx
i 1
1
4
As n increases, the value of Sn increases, getting
closer and closer to the true are under the curve.
Integral Notation
We write the value of the midpoint sum as n
gets very large by an integral
b
 f ( x) dx  Area under f ( x) between x  a and x  b
a
Find Revenue from Buffalo Dinners Not Sold:
Optimum price and quantity are $19.19 and 2300
$32
$24
$19.99
D(q) = 0.0000018q2  0.0002953q + 30.19
Consumer
Surplus
Demand Function
$16
q = 2,300
$8
D(2,300) = $19.99
Revenue
$45,977
Not
Sold
4014
$0
0
1000
2000
3000
2,300
4,014
Not 
Sold
 D(q) dq  $18,643
2,300
4000
5000
Find Consumer Surplus for Dinners
$32
$24
$19.99
D(q) = 0.0000018q2  0.0002953q + 30.19
Consumer
Surplus
Demand Function
$16
q = 2,300
$8
D(2,300) = $19.99
Revenue
$45,977
Not
Sold
$0
0
1000
2000
3000
4000
2,300
Consumer

Surplus
2,300
 D(q) dq  R(q) $61,356  $45,977  $15,379.
0
5000