Topic 10:
Integration
Jacques
Indefinate Integration 6.1
Definate Integration 6.2
1
Intuition
y = F (x) = xn + c
dy/dx = F`(x) = f(x) = n x
n-1
Given the derivative f(x), what is
F(x) ? (Integral, Anti-derivative or
the Primitive function). The
process of
integration.
finding
F(x)
is
2
Definition
Just as f(x) = derivative of F(x)
F ( x)   f ( x)dx
Example
F ( x) 
2
3
3
x
dx

x
c

c=constant of integration (since derivative
of c=0)of course, c may be =0….., but it
may not
check: if y = x3 + c then dy/dx = 3x2
or if c=0, so y = x3 then dy/dx = 3x2
3
Rule 1 of Integration:
1
x n 1  c
F ( x)   x dx 
n 1
n
1 3
F ( x)   x dx  x  c
3
2
1
3
check: if y = /3 x + c then dy/dx = x
2
F ( x )   dx   1.dx   x dx  x  c
0
check: if y = x + c then dy/dx = 1
4
Rule 2 of Integration:
F ( x)   af ( x)dx  a  f ( x)dx
Examples
1 
F ( x )   3x 2 dx  3 x 2 dx  3. .x3   c  x3  c
3 
check…..
F( x )   a.dx  a dx ax  c
check…
F( x )   4dx  4 dx 4x  c
check
5
• Rule 3 of Integration:
F ( x)    f ( x)  g ( x)dx   f ( x)dx   g ( x)dx
• Example
F ( x)   3x
2
 2 xdx   3x dx   2 x dx  x
2
3
 x c
2
6
Calculating Marginal Functions
d TR 
MR 
dQ
d TC 
MC 
dQ
•Given MR and MC use integration
to find TR and TC
TRQ   MRQ.dQ
TCQ   MC Q.dQ
7
Marginal Cost Function
Given the Marginal Cost Function, derive
an expression for Total Cost?
MC = f (Q) = a + bQ + cQ2


TC ( Q )   a  bQ  cQ 2 dQ
TC ( Q )  a  dQ  b  Q dQ  c  Q 2 dQ
b 2 c 3
TC ( Q )  aQ  Q  Q  F
2
3
F = the constant of integration
If Q=0, then TC=F
F= Fixed Cost….. (or TC when Q=0)
8
Another Example
MC = f (Q) = Q + 5
Find an expression for Total Cost in
terms of Q, if TC = 20 when
production is zero.
TC( Q )   Q  5dQ
TC( Q )   Q dQ  5 dQ
TC ( Q ) 
1 2
Q  5Q  F
2
F = the constant of integration
If Q=0, then TC = F = Fixed Cost
So if TC = 20 when Q=0, then F=20
So,
1 2
TC ( Q )  Q  5Q  20
2
9
Another Example
Given
Marginal
Revenue,
MR = f (Q) = 20 – 2Q
Find the Total Revenue function?
MR = f (Q) = 20 – 2Q
TR( Q )   20  2QdQ
TR( Q )  20 dQ  2 QdQ
TR( Q )  20Q  Q 2  c
c = the constant of integration
10
Example:
Given MC=2Q2 – 6Q + 6; MR = 22 – 2Q;
and Fixed Cost =0. Find total profit for profit
maximising firm when MR=MC?
1) Find profit max output Q where MR = MC
MR=MC
2
so 22 – 2Q = 2Q – 6Q + 6
2
gives Q – 2Q – 8 = 0
(Q - 4)(Q + 8) = 0 so Q = +4 or Q =-2
Q = +4
11
2) Find TR and TC
TR( Q )   22  2QdQ
TR( Q )  22 dQ  2 QdQ
TR( Q )  22Q  Q 2  c
so TR = 22Q – Q2
MC = f (Q) = 2Q2 – 6Q + 6
TC ( Q ) 


2
 2Q  6Q  6 dQ
TC ( Q )  2  Q 2dQ  6  QdQ  6  dQ
2 3
Q  3Q 2  6Q  F
3
F = Fixed Cost = 0
(from question)
2 3
TC ( Q )  Q  3Q 2  6Q
so….
3
TC ( Q ) 
12
3. Find profit = TR-TC, by substituting in
value of q* when MR = MC
Profit = TR – TC
2
TR if q*=4: 22(4) - 4 = 88-16 = 72
TC if q* =4: 2/3 (4)3 – 3(4)2 + 6(4) = 2/3(64)
– 48 + 24 = 182/3
Total profit when producing at MR=MC so
q*=4 is
2
1
TR – TC = 72 - 18 /3 = 53 /3
13
Some general points for answering
these types of questions
Given a MR and MC curves
- can find profit maximising output q* where
MR = MC
- can find TR and TC by integrating MR
and MC
- substitute in value q* into TR and TC to
find a value for TR and TC. then…..
- since profit = TR – TC
can find (i) profit if given value for F or (ii)
F if given value for profit
14
Definite Integration
The definite integral of f(x) between
values a and b is:
F ( x ) 
b
b
a

 f ( x ) dx  F (b )  F ( a )
a
15
Example
2
1
1
7
1 3 
3
3
1 x dx   3 x 1  3 (2)  3 (1)  3
2
2
1)
6
2)
 3dx  3x
6
2
 3(6)  3( 2)  12
2
16
Definition
b
f ( x )dx

The definite integral a
can be
interpreted as the area bounded by the
graph of f(x), the x-axis, and vertical
lines x=a and x=b
f(x)
a
b
x
17
Consumer Surplus
Difference between value to consumers and to the
market…. Represented by the area under the
Demand curve and over the Price line…..
P
P1
0
x
Demand Curve:
P = f(Q)
a
Q1
Consumer Surplus
Q
18
Or more formally….
CS(Q) = oQ1ax - oQ1aP1
Where oQ1ax represents the entire area under the
demand curve up to Q1 and
oQ1aP1 represents the area in the rectangle, under the
price line up to Q1
Hence,
Q1
CS (Q )   D (Q )dQ  P1Q1
0
19
Producer Surplus
Difference between market value and total cost
to producers…. Represented by the area over
the Supply curve and under the Price line…..
P
Supply Curve:
P = g(Q)
P1
y
0
a
Producer Surplus
Q1
Q
20
Or more formally….
PS(Q) = oQ1aP1 - oQ1ay
Where oQ1aP1 represents the area of the entire
rectangle under the price line up to Q1 and
oQ1ay represents the area under the Supply curve up
to Q1
Hence
Q1
PS ( Q )  P1Q1   S ( Q )dQ
0
21
Example 1…..
Find a measure of consumer surplus
at Q = 5, for the demand function P =
30 – 4Q
Solution
1) solve for P at Q = 5
If Q = 5, then P = 30 – 4(5) = 10
22
The picture….
2) ‘sketch’ diagram
P = 30 – 4Q
intercepts: (0, 30) and (7.5, 0)
At Q = 5, we have P = 10 ….. Draw in price line….
P
30
Demand Curve:
P = f(Q) = 30 – 4Q
Consumer Surplus
P1=10
0
Q1 = 5
7.5
Q
23
Calculation…
Q1
 D(Q)dQ  P Q
3) Evaluate Consumer Surplus
CS (Q) 
1
1
0
i) Entire area under demand curve between 0 and
Q1= 5:
 (30  4Q)dQ  30Q  2Q 
5
2 5
0
0
 30(5)  2( 25)  0  100
ii) total revenue = area under price line at P1 = 10,
between Q = 0 and Q1 = 5 is P1Q1 = 50
iii) So CS = 100 – p1Q1 = 100 – (10*5) = 50
24
Example 2
If p = 3 + Q2 is the supply curve, find a
measure of producer surplus at Q = 4
Solution
1) evaluate P at Q = 4
If Q = 4, then p = 3 + 16 = 19
25
The picture….
2) ‘Sketch’ the diagram
P = 3 + Q2
intercept: (0, 3)
Price line at Q = 4, P = 19
P
Supply Curve:
P = g(Q) = 3 + Q2
P1 = 19
Producer Surplus
3
0
Q1 = 4
Q
26
Calculation…
PS ( Q )  P1Q1   S ( Q )dQ
Q1
0
3) Evaluate Producers Surplus
i) Entire area under supply curve between Q = 0 and Q1 = 4…..
4
4
1


2
3
(
3

Q
)
dQ

3
Q

Q
0

3  0

1
 3(4)  (4) 3  0  33 13
3
ii) total revenue = area under price line (p1 = 19), between Q = 0
and Q1 = 4 , and this = p1Q1 = 76
iii) So PS = p1Q1 – 331/3 =
76 – 331/3 = 422/3
27
Example 3
• The inverse demand and supply functions
for a good are, respectively:
•
and
P  2Q  14
P Q2
• Find the market equilibrium values of P
and Q.
• Find the Total surplus (CS + PS) when the
market is in equilibrium.
28
Find market equilibrium….
At equilibrium
2Q  14  Q  2
3Q  12
So equilibrium Q*  4
Thus equilibrium P*  4  2  6
29
‘sketch’ the diagram
P
CS
14
S
14
P*=6
PS
2
0
Consumer Surplus
D
Q* = 4
7
Q
30
Consumer surplus…
CS 

Q*
0
DQ dQ  P * Q *
i) area under entire demand curve between Q = 0 and
Q*
   2Q  14 dQ
4
0


4
  Q 2  14Q 0

 

  4   144    0   140 
2
2
 16  56  40
ii) total revenue = area under price line at P* = 6, between
Q = 0 and Q* = 4 is P*Q* = 24
iii) So CS = 40 – 24 = 16
31
Producer Surplus…
PS  P Q  
*
*
Q*
0
S Q .dQ
i) area under Supply curve between Q = 0 and Q*
  Q  2dQ
4
0
4
1

  Q 2  2Q 
2
0
 1 2
 1 2

  4  24   0  20
 2

 2
 8  8  16
ii) total revenue = area under price line at P* = 6, between Q = 0
and Q* = 4 is P*Q* = 24
iii) So PS = 24 – 16 = 8
32
Total Surplus
• Total surplus = CS + PS = 16 + 8
= 24
33