T10

Integration

Course Manual

Indefinite Integration 7.1-7.2

Definite Integration 7.3-7.4

Jacques (3rd Edition)

Indefinite Integration 6.1

Definite Integration 6.2



n

y = F (x) = x + c

n-1

dy/dx = F`(x) = f(x) = n x



Given the derivative f(x), what is

F(x) ? (Integral, Anti-derivative

or the Primitive function).

Just as f(x) = derivative of F(x)



F ( x)   f ( x)dx



Example



F ( x)   3x dx  x  c

2 3







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



How did we integrate f(x)?

Rule 1 of Integration:



1 n1

F ( x)   x dx 

n

x c

n 1



Examples

1 3

F ( x)   x dx  x  c

2



3

check: if y = 1/3 x3 + c then dy/dx = x2





F ( x )   dx   1.dx   x 0 dx  x  c

check: if y = x + c then dy/dx = 1

Rule 2 of Integration:



F ( x)   af ( x)dx  a  f ( x)dx



Examples

1 3

F ( x )   3x dx  3 x dx  3. .x   c  x3  c

2 2

3 



F ( x )   a.dx  a  dx ax  c



F ( x )   4dx  4 dx 4 x  c

Rule 3 of Integration:



F ( x)    f ( x)  g ( x)dx   f ( x)dx   g ( x)dx



Example

 

F ( x)   3x2  2 x dx   3x2dx   2 x dx  x3  x2  c

•Calculating Marginal Functions



d TR  d TC 

MR  MC 

dQ dQ







•Given MR and MC use integration

to find TR and TC

TRQ   MRQ.dQ



TCQ   MCQ.dQ

Marginal Cost Function

Given the Marginal Cost Function, derive

an expression for Total Cost?

2

MC = f (Q) = a + bQ + cQ



 

TC ( Q )   a  bQ  cQ 2 dQ





TC ( Q )  a  dQ  b  Q dQ  c  Q dQ

2







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…..

Another Example



MC = f (Q) = Q + 5

If Total Cost = 20 when production

is 0, find TC function?



TC( Q )   Q  5dQ

TC( Q )   Q dQ  5 dQ

1 2

TC ( Q )  Q  5Q  F

2

F = the constant of integration

If Q=0, then TC = F = Fixed Cost

So if TC = 20 then,

1 2

TC ( Q )  Q  5Q  20

2

Another Example

Given Marginal Revenue, 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

Example:



Given MC=2Q2 – 6Q + 6; MR = 22 – 2Q;

and Fixed Cost =0. Find total profit for profit

maximising firm when MR=MC?



Solution:

1) Find profit max output Q where MR = MC

MR=MC

so 22 – 2Q = 2Q2 – 6Q + 6

gives Q2 – 2Q – 8 = 0

(Q - 4)(Q + 8) = 0 so Q = +4 or Q =-2

Q = +4



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 )   2Q 2  6Q  6 dQ

TC ( Q )  2  Q 2dQ  6 QdQ  6  dQ

2 3

TC ( Q )  Q  3Q 2  6Q  F

3

F = Fixed Cost = 0 (from question)

2 3

TC ( Q )  Q  3Q 2  6Q

so…. 3



3. Find profit = TR-TC, by substituting in

value of q* when MR = MC



Profit = TR – TC

TR if q*=4: 22(4) - 42 = 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

TR – TC = 72 - 182/3 = 53 1/3

NOTE:

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

Definite Integration

The definite integral of f(x) between

values a and b is:

b

F ( x)   f ( x)dx  F (b)  F (a)

b

a

a



Example

2

1 3 

2

1 3 1 3 7

 x dx   3 x 1  3 (2)  3 (1)  3

2



1  







6



 3dx  3x2  3(6)  3(2)  12

6



2

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

The Consumer Surplus

Difference between value to consumers

and to the market….

P

x Demand Curve:

P = f(Q)



Consumer Surplus

a

P1









0 Q1 Q





CS(Q) = oQ1ax - oQ1aP1

Q1



CS (Q)   D(Q)dQ  P Q1

1

0

Producer Surplus

Difference between market value and total

cost to producers…



P

Supply Curve:

P = g(Q)



P1 a

Producer Surplus

y







0 Q1 Q





PS(Q) = oQ1aP1 - oQ1ay

Q1

PS ( Q )  P Q1   S ( Q )dQ

1

0

examples…..

Find a measure of consumer surplus

at Q = 5,

for the demand function p = 30 – 4Q



Solution

If Q = 5, then p = 30 – 4(5) = 10

Q1



CS (Q)   D(Q)dQ  P Q1

1

0







P

30 Demand Curve:

P = f(Q) = 30 – 4Q



Consumer Surplus

P1=10









0 Q1 = 5 7.5

Q

Entire area under demand curve between 0

and Q1 = 5:

5

 ( 30  4Q )dQ   25

30Q  2Q 0 

0

 30( 5 )  2( 25 )  0  100





total revenue = area under price line

(p1 = 10), between Q = 0 and Q1 = 5 is

p1Q1



So CS = 100 – p1Q1 = 100 – (10*5) = 50

Example 2:



If p = 3 + Q2 is the supply curve, find a

measure of producer surplus at Q = 4



Solution

If Q = 4, then p = 3 + 16 = 19



Q1

PS ( Q )  P Q1   S ( Q )dQ

1

0





P

Supply Curve:

P = g(Q) = 3 + Q2





P1 = 19 Producer Surplus



3







0 Q1 = 4

Q

Entire area under supply curve between

Q = 0 and Q1 = 4…..

4 4

 1 

 ( 3  Q 2 )dQ  3Q  Q 3 

0  3 0

1

 3( 4 )  ( 4 )3  0  33 1

3

3



total revenue = area under price line

(p1 = 19), between Q = 0 and Q1 = 4 is

p1Q1 = 76



So PS = p1Q1 – 331/3 =

76 – 331/3 = 422/3

Manual, Topic 7

Q3. A profit maximising firm has MR  34  3Q

and MC  Q 2  10Q  26 . How much will it

produce? What level of fixed costs would

make the firm make zero profits?



Step 1: set MR=MC and find output that

maximises profit, q*

Q 2  10Q  26  34  3Q

Q 2  7Q  8  0



Solve the quadratic for value of Q using

 b  b 2  4a c 

Q

formula 2a  :



a=1, b=-7, c=-8

7  49  41 8 7  9

Q 

so

21 2

Q  1 (inadmissible) or Q  8

Thus 8 units produced by profit max firm

Step 2: integrate MR and MC to find TR &

TC, and thus profits



  TR  TC

TR   MR.dQ   34  3Q dQ  34Q  Q 2  c

3

2

In this case, the constant of integration c  0 ,

since the firm makes no revenue when Q=0



  1

TC   MC.dQ   Q 2  10Q  26 dQ  Q 3  5Q 2  26Q  F

3



F, the constant of integration = Fixed Costs

3 2 1 3

  34Q  Q  Q  5Q 2  26Q  F

2 3

1 3 7 2

   Q  Q  8Q  F

3 2

Step 3: substitute in q* to TR and TC to get

profit max values when producing q*



Substituting in Q  8 for profit max.



  8  8  88  170 2  224  64  117 1  F

1 3 7 2

3 2 3 3



Step 4: Set profit =0 (thus TR – TC = 0), &

solve for F



1

Setting   0 , gives 0  117  F

3

1

Thus, value of F at =0 is F  117

3

Q4 (b): A firm which has no fixed costs has

MC and MR given as follows:

MC=2Q2 – 6Q + 6;

MR = 22 – 2Q;

Find total profit for profit maximising firm

when MR=MC?



Solution:

1) Find profit max output Q where MR = MC



22 – 2Q = 2Q2 – 6Q + 6

gives Q2 – 2Q – 8 = 0

Solve quadratic for Q, by using formula, or

(Q - 4)(Q + 8) = 0 so Q = +4 or Q =-2

so Q = +4 (since Q=-2 inadmissable)



2) Find TR and TC

TR( Q )   22  2Q dQ

TR( Q )  22 dQ  2 QdQ

TR( Q )  22Q  Q 2  c

TR = c when Q=0; but TR = 0 when Q = 0; so

therefore c = 0



so TR = 22Q – Q2



MC = f (Q) = 2Q2 – 6Q + 6

 

TC ( Q )   2Q 2  6Q  6 dQ



TC ( Q )  2  Q 2dQ  6 QdQ  6  dQ

2 3

TC ( Q )  Q  3Q 2  6Q  F

3

F = Fixed Cost = 0 (from question)

2 3

TC ( Q )  Q  3Q 2  6Q

so…. 3

3. Find profit = TR-TC, by substituting in

value of q* when MR = MC



Profit = TR – TC

TR if q*=4: 22(4) - 42 = 88-16 = 72

2

TC if q* =4: /3 (4)3 – 3(4)2 + 6(4)

= 2/3(64) – 48 + 24

= 182/3

so total profit when producing at MR=MC at

q*=4 is

TR – TC = 72 - 182/3 = 53 1/3

Q5. The demand and supply functions for

a good are given by the equations P  2Q  14

and P  Q  2 respectively. Determine the

equilibrium price and quantity and

calculate the consumer and producer

surplus at equilibrium.



At equilibrium

2Q  14  Q  2

3Q  12

So equilibrium Q*  4

Thus equilibrium P*  4  2  6



P CS

14 14

S





P*=6



PS 2 D



0 Q* = 4 7

Q

Consumer Surplus









Consumer Surplus

Difference between value to consumers and

to the market…. Area above price line and

under Demand curve





D Q dQ  P * Q *

Q*

CS  0







  2Q  14dQ  64

4

CS 

0







CS   Q  14Q  24

2

4

0





   

CS   42  144   02  140  24



CS  16  56  24  16

Producer Surplus

Difference between market value and total

cost to producers… area below price line and

above Supply curve





PS  P1Q1  0 S Q .dQ

Q1







PS  64  0 Q  2dQ

4





4

1 

PS  24   Q 2  2Q 

2 0



 1  1 

PS  24   42  24   02  20

 2  2 



PS  24  8  8  8

Total Surplus = CS + PS = 16 + 8 = 24