Calculus Lecture

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Department of Mathematics

London School of Economics and Political Science

MA 100 – Mathematical Methods

Calculus – Lecture 16

Consumers’ and Producers’ Surplus

Integration by Partial Fractions,

by Change of Variable, and by Parts

Department of Mathematics

London School of Economics and Political Science

First an application : Economic model of a market

Consider a market with one product.

Suppose the inverse demand function is

p = D(x) = 60 − 3 x ( 0 ≤ x ≤ 20 );

i.e., in order to sell x units of the product,

the price per unit must be p = D(x) .

And the inverse supply function is p = S(x) = x2 + 20,

i.e., for it to be worthwhile to produce x units,

the price per unit must be p = S(x) .

MA 100, Mathematical Methods – Calculus – Lecture 16 – page 2 / 42

The Market at Equilibrium

The inverse supply and demand functions meet

at a point (x0, p0) .

The value of p0 is the market price.

To find this point, we must solve : D(x) = S(x) .



x

p

x0

p0

D(x)

S(x)



At the market price p0 , the total the consumers pay

and the total the producers receive is :

R = x0 p0 .

This is the area of the shaded region below.

x0

p0


Consumers’ Surplus

At the market price p0 , consumers who buy the product

would have been willing to pay at least p0 , and many of

them would have been willing to pay more.

This extra in the market forms the Consumers’ Surplus.

It is the area of the shaded region below.

x0

p0

D(x)

S(x)


Consumers’ Surplus

The value of the consumers’ surplus, the area of that

region, is given by the definite integral

∫ x0

x=0

D(x) dx − x0 p0 =

∫ x0

x=0

(D(x) − p0) dx .

In this case, the integral is equal to∫ 5

x=0

(60 − 3 x − 45) dx =


Producers’ Surplus

Similarly, the Producers’ Surplus is the area of the region

x0

p0

D(x)

S(x)

This is obtained by determining

x0 p0 −

∫ x0

x=0

S(x) dx =

∫ x0

x=0

(p0 − S(x)) dx .


Techniques of Integration

In these examples, integrating the functions is quite

straightforward.

You are expected to be familiar with

“standard” integrals,

and “standard” methods of integration.

If you could use a refresher, see ( for instance ) :

the Calculus Refresher on the MA100 Moodle page,

or the relevant sections of the Calculus textbook.

The rest of the lecture is devoted to some more advanced

techniques of integration.


Integration – a quick reminder

Some standard integrals :

f (x)

∫

f (x) dx

xn ( n 6= −1 )1

n + 1xn+1 + c

1/x ln |x | + c

ex ex + c

sin(x) −cos(x) + c

cos(x) sin(x) + c


Some further integrals

Some other integrals we know by now :∫

dx√

1 − x2= arcsin(x) + c ;

∫

dx

1 + x2= arctan(x) + c.

Why ?

Becaused

dxarcsin(x) =

1√

1 − x2, etc.


Partial fractions – Example

Find :

∫

x + 2

x2 + 5 x + 4dx .

We have

x + 2

x2 + 5 x + 4=

x + 2

(x + 1) (x + 4)=

1/3

x + 1+

2/3

x + 4,

so :

∫

x + 2

x2 + 5 x + 4dx =

1

3

∫

dx

x + 1+

2

3

∫

dx

x + 4

=1

3ln|x + 1| +

2

3ln|x + 4| + c.


Integration – Rational Functions

We first look at the integration of rational functions :

functions of the formP(x)

Q(x),

with P(x) and Q(x) polynomials.

Example

∫

x5 + 25

2 x4 + 4 x3 + 10 x2dx .


Integration – Rational Functions

If the degree of P is greater than or equal to that of Q ,

then we can always write :P(x)

Q(x)= D(x) +

R(x)

Q(x),

with D , R , Q all polynomials,

and where the degree of R is smaller than that of Q .

Next : if Q(x) = a xk + b xk−1 + · · · ( with a 6= 0 ),

so Q(x) = a(

xk +b

axk−1 + · · ·

)

,

then we can write :R(x)

Q(x)=

1

a·

R(x)

xk + ba

xk−1 + · · ·.


Integration – rational functions

So we can reduce integrating any rational function

to the case

∫

P(x)

Q(x)dx =

∫

[

D(x) +1

a·

R(x)

xk + · · ·

]

dx

=

∫

D(x) dx +1

a

∫

R(x)

xk + · · ·dx ;

where R(x) has degree smaller than k .


Partial fractions – the easy case

Suppose we have a function of the form

R(x)

xk + · · ·=

R(x)

(x − a1) (x − a2) · · · (x − ak),

where R(x) is a polynomial of degree smaller than k ,

and a1, a2, . . . , ak are different real numbers.

Such a function can always be written in the form

R(x)

xk + · · ·=

A1

x − a1

+A2

x − a2

+ · · · +Ak

x − ak

,

for some constants A1, A2, . . . ,Ak .


Partial fractions – finding the constants

But how can we find the A1,A2, . . . ,Ak ?

Method 1 : just work it out !

Example

x2 + 3

(x − 1) (x + 1) (x − 2)=

A

x − 1+

B

x + 1+

C

x − 2

=A (x + 1) (x − 2) + B (x − 1) (x − 2) + C (x − 1) (x + 1)

(x − 1) (x + 1) (x − 2).


Partial fractions – finding the constants

So we must have :

x2 + 3 = 1 x2 + 0 x + 3

= (A + B + C ) x2 + (−A − 3 B ) x + (−2 A + 2 B − C );

which means we have to solve :

A + B + C = 1,

−A − 3 B = 0,

−2 A + 2 B − C = 3.

The solution is : A = , B = , C =


Partial fractions – the cover-up rule

Method 2 : the cover-up rule

The start is the same :

x2 + 3

(x − 1) (x + 1) (x − 2)=

A

x − 1+

B

x + 1+

C

x − 2

=A (x + 1) (x − 2) + B (x − 1) (x − 2) + C (x − 1) (x + 1)

(x − 1) (x + 1) (x − 2).

So we need x2 + 3 = A (x + 1) (x − 2)

to make sure :+ B (x − 1) (x − 2)

+ C (x − 1) (x + 1) .



To find A :

The constant A came from the fractionA

x − 1.

Set x = 1 everywhere in the equation :

12 + 3 = A (1 + 1) (1 − 2) + B · 0 + C · 0.

Because of the zeros in the terms without an A ,

we can describe this as :

“cover up the terms with the factor x − 1 on the right”.

This gives an equation for A only :

Which easily gives :



We can do the same for B ( coming fromB

x + 1) :

take x = −1 : (−1)2 + 3 = B (−1 − 1) (−1 − 2) .

So :

And then for C ( coming fromC

x − 2) :

take x = 2 : 22 + 3 = C (2 − 1) (2 + 1) .

This gives :


Partial fractions – the result

So now we can integrate :

∫

x2 + 3

(x − 1) (x + 1) (x − 2)dx

=

∫

[ −2

x − 1+

2/3

x + 1+

7/3

x − 2

]

dx

= −2 ln|x − 1| +2

3ln|x + 1| +

7

3ln|x − 2| + c.


Partial fractions – with repeated factors

When there are repeated factors, more work is needed.

Example :

∫

x − 1

x (x + 1)2dx .

Write :x − 1

x (x + 1)2=

A

x+

B

x + 1+

C

(x + 1)2

=A (x + 1)2 + B x (x + 1) + C x

x (x + 1)2.

So we need to make sure that A , B , C satisfy :

x − 1 = A (x + 1)2 + B x (x + 1) + C x .



Apply cover-up rule, taking x = 0 : 0 − 1 = A (0 + 1)2 ,

which gives :

Now take x = −1 : (−1) − 1 = C (−1) ,

hence :

But we can’t get B that way.

Use the first method :

x − 1 = (A + B) x2 + (2 A + B + C) x + A .

So A + B = 0, hence :



Alternative approach to finding B :

So far we have :

x − 1

x (x + 1)2=

−1

x+

B

x + 1+

2

(x + 1)2.

Now evaluate both sides at some convenient value for x ,

say at x = 1.



So now we can integrate :

∫

x − 1

x (x + 1)2dx =

∫

[

−1

x+

1

x + 1+

2

(x + 1)2

]

dx

= −ln|x | + ln|x + 1| −2

x + 1+ c .


Integration – the easy case in general

If we can completely factorise Q(x) = xk + · · · :

Q(x) = (x − a1) (x − a2) · · · (x − ak) ,

with all a1, a2, . . . , ak different,

then we can write

R(x)

Q(x)=

A1

x − a1

+A2

x − a2

+ · · · +Ak

x − ak

.

And those fractions are easy to integrate :∫

Ai

x − ai

dx = Ai ln |x − ai | + c .


Partial fractions – with repeated factors in general

If there are repeated factors, e.g.:

Q(x) = (x − a1)3 (x − a2) (x − a3)

2 ,

then we can write

R(x)

Q(x)=

A

x − a1

+B

(x − a1)2+

C

(x − a1)3

+D

x − a2

+E

x − a3

+F

(x − a3)2.


Partial fractions – with repeated factors in general

These functions can be integrated :

∫

A

x − a1

dx = A ln |x − a1| + c ,

∫

B

(x − a1)2dx = −

B

x − a1

+ c′,

∫

C

(x − a1)3dx = −

1

2·

C

(x − a1)2+ c′′,

etc.


Partial fractions – with irreducible squares

We can’t factorise, say, x2 + 2 x + 5

as (x − a1) (x − a2) ( it is irreducible ).

So what if those appear in the rational function ?

We can always complete the square :

x2 + a x + b =(

x + 12

a)2

+(

b − 14

a2)

,

where b − 14

a2 > 0. ( Otherwise we could factorise ! )



Now combine this with two useful observations :

d

dxarctan

(x + p

q

)

=1

q·

1

1 +(x + p

q

)2

=q

x2 + 2 p x + (p 2 + q 2);

and

d

dxln(x2 + a x + b) =

2 x + a

x2 + a x + b.



Example

To find

∫

2 x − 1

x2 + 2 x + 5dx , we should use

d

dxln(x2 + 2 x + 5) =

2 x + 2

x2 + 2 x + 5,

and

d

dxarctan

(x + 1

2

)

=2

(x + 1)2 + 22=

2

x2 + 2 x + 5.



So we find :

∫

2 x − 1

x2 + 2 x + 5dx

=

∫

2 x + 2

x2 + 2 x + 5dx +

−3

2

∫

2

x2 + 2 x + 5dx

= ln(x2 + 2 x + 5) −3

2arctan

(x + 1

2

)

+ c.


Partial fractions – all together now

We may need to combine all these approaches; e.g., to find

∫

x5 + 25

2 x4 + 4 x3 + 10 x2dx .

First rewrite :

x5 + 25

2 x4 + 4 x3 + 10 x2=

We can factorise ( a bit ) :

x4 + 2 x3 + 5 x2 =


Partial fractions – all together now

So now we need to look for A , B , C , D so that

12(−x3 + 10 x2 + 25)

x4 + 2 x3 + 5 x2=

And continue by integrating each factor one by one

( and not forgetting the terms 12

x − 1 ).


Integration by change of variables ( = by substitution )

This technique is based on the Chain Rule :

d

dxf (g(x)) = f ′(g(x)) g′(x) .

So we have :

∫

f ′(g(x)) g′(x) dx = f (g(x)) + c .

General idea :

Whenever you see a composite function f (g(x)) ,

try the substitution u = g(x) .


Integration by change of variables

Example :

∫

x ex2+1 dx .

Try to substitute : u = x2 + 1;

then :du

dx= 2 x , and so : “x dx = 1

2du ”.

So we find :

∫

x ex2+1 dx =

∫

eu 12

du =


Integration by change of variables

Example :

∫

x (3 x + 5)7 dx .

Try : u = 3 x + 5, hence : x = 13(u − 5) ;

then alsodu

dx= 3, and so : “dx = 1

3du ”.

We find :∫

x (3 x + 5)7 dx =

∫

13(u − 5) u7 1

3du =


Integration by change of variables – definite integrals

Be careful with definite integrals :

∫ b

a

f ′(g(x)) g′(x) dx = f (g(b)) − f (g(a)) =

∫ g(b)

g(a)f ′(u) du .

Example :

∫ 1

x=0

x ex2+1 dx .

Using the substitution u = x2 + 1 again, this becomes :∫ 1

x=0

x ex2+1 dx =

∫

eu 12

du =


Integration by parts

This technique is based on the Product Rule :

d

dxf (x) g(x) = f ′(x) g(x) + f (x) g′(x) .

So we have :

f (x) g(x) =

∫

f ′(x) g(x) dx +

∫

f (x) g′(x) dx .

Hence :

∫

f ′(x) g(x) dx = f (x) g(x) −

∫

f (x) g′(x) dx ;

and possibly / hopefully : f (x) g′(x) is easier to integrate.



Example :

∫

x ln(x) dx .

Try : f ′(x) = x , so :

and : g(x) = ln(x) , so :

So we find :∫

x ln(x) dx =



A trick that is useful sometimes is :

∫

f (x) dx =

∫

1 f (x) dx = x f (x) −

∫

x f ′(x) dx .

Example :

∫

arctan(x) dx = x arctan(x) −

∫

x1

1 + x2dx =


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