As the students of M.B.A, we need to know about many things regarding business mathematics. An entrepreneur needs to take many decisions with the help of mathematical terms. By using the terms properly, an entrepreneur can easily take any decision quantitatively. Among many tools, Integral calculus is one of them. It helps to identify many things related to our practical business life. Such as: total cost, total revenue, producer’s surplus, consumer’s surplus etc. So, it’s an important element of business tools. This report is completely based on integral calculus and it’s uses in different fields.

1. Introduction
As the students of M.B.A, we need to know about many things regarding
business mathematics. An entrepreneur needs to take many decisions
with the help of mathematical terms. By using the terms properly, an
entrepreneur can easily take any decision quantitatively. Among many
tools, Integral calculus is one of them. It helps to identify many things
related to our practical business life. Such as: total cost, total revenue,
producer’s surplus, consumer’s surplus etc. So, it’s an important element
of business tools. This report is completely based on integral calculus
and it’s uses in different fields.
Objectives
The main objectives of this report are:
1. To know about integration in details.
2. To know the methods and techniques of integration.
3. To know the uses of integral calculus in our day to day life.

2. Integration
Integration from the Latin integer meaning whole or entire generally
means combining parts so that they work together or form a whole.
Integration is often introduced as the reverse process to differentiation,
and has wide applications, for example in finding areas under curves.
Process of Integration:
Types of Integration: There are basic two types of integration. They
are:
ECTIVES

3. Definite Integral:
Techniques of Integration: There are some major techniques of
integration are:
Integration as summation
Integration may be introduced as a means of finding areas using
summation and limits
Integration using a table of anti-derivatives
Integration may be regarded as the reverse of differentiation, so a table
of derivatives can be read backwards as a table of anti-derivatives.
Integration by parts
A special rule, integration by parts, can often be used to integrate the
product of two functions. It is appropriate when one of the functions
forming the product is recognised as the derivative of another function.
The result still involves an integral, but in many cases the new integral
will be simpler than the original one.
Integration by substitution
There are occasions when it is possible to perform an apparently difficult
piece of integration by first making a substitution. This has the effect of
changing the variable and the integrand. With definite integrals the
limits of integration can also change.

4. Application Of Integration:
· define Total Cost, Variable Cost, Average Cost, Marginal Cost,
Total Revenue, Marginal Revenue and Average Revenue;
· find marginal cost and average cost when total cost is given;
· find marginal revenue and average revenue when total revenue is
given;
· find total cost/ total revenue when marginal cost/marginal
revenue are given, under given conditions.
. define and calculate a consumer’s surplus
. define and calculate a producer’s surplus.
Determination of cost function:
If C denotes the total cost and MC is the marginal cost, then we can
write
C Cx MCdx k , where k is the constant of integration, k, being the
constant, is the fixed cost.
Example: Given
MC = 5+16x-3x2
C(x) =  (5+16x-3x2)dx
C(x) = 5x +8x2 -x3 +k
When x = 5, C(x) = C(5) = Rs. 500
or, 500= 25+200-125+ k

5. This gives, k = 400
C(x)= 5x +8x2 -x3 +400
Determination of Total Revenue Function:
If R(x) denotes the total revenue function and MR is the marginal
revenue function, then
R(x)=  (MR)dx+k Where k is the constant of integration.
R(X)
Also, when R (x) is known, the demand function can be found as
p= X
Example:
The marginal revenue function of a commodity is given as
MR =12-3x2 +4x . Find the total revenue function.
MR = 12-3x2 +4x
R =  (12-3x2 +4x)dx+k
R =12x-x3 +2x2 [constant of integration is zero in this case]
Revenue function is given by R = 12x +2x2-x3

6. Average Value of a Function Over an Interval:
EXAMPLE:
During a certain 12-hour period the temperature at time
t (measured in hours from the start of the period) was
degrees. What was the average temperature during that period?
SOLUTION:
The average temperature during the 12-hour period from t = 0 to t
= 12 is
2
3
1
447 tt −+
12
0
3
2
12
0
2
9
247
12
1
3
1
447
012
1






−+=





−+
− ∫
t
ttdttt
( ) ( ) ( ) ( ) 











−+−





−+=
9
0
02047
9
12
1221247
12
1 3
2
3
2
[ ] .degrees550660
12
1
=−=

7. Consumer surplus:
The consumer surplus represents the total savings to consumers
who are willing to pay more than price for the product.
The consumer’s surplus at a price level of p is
( )
0
Consumer surplus (CS)
q
D x dx pq= −∫
Where ( )D x is demand function and pq is the revenue.
EXAMPLE:
Find the consumers’ surplus for the following demand curve at the given
sales level x. 20;
10
3 =−= x
x
p
SOLUTION:
Therefore, the consumers’ surplus is
dx
x
dx
x
∫∫ 





−=





−





−
20
0
20
0 10
21
10
3
20
0
2
20
2
x
x −=
( ) ( ) 





−−





−=
20
0
02
20
20
202
22
[ ] [ ] .20020 =−=
That is, the consumers’ surplus is $20.
Graphical Presentation of Consumers’ Surplus.
.123
10
20
3 =−=−=P

8. • On the graph, the consumer surplus (yellow) is the area located
below the Demand function and above the rectangle that
represents the revenue generated (red).

9. Supplier’s/ Producer’s surplus: The supplier’s surplus measures
the difference between the amount of money a supplier is willing to
accept at a given price for a product and the amount the supplier actually
does receive.
The supplier’s surplus at a price level of p is
( )
0
Supplier surplus (SS)
q
pq S x dx= − ∫
Where ( )S x is supply function and pq is the revenue.
EXAMPLE:
At market equilibrium , consumers demand 100 (000) tons if SAE 90
lubricating oil, whose supply function is Ps (q) = 10+0.5 q
Where q is the thousands tons and ps(q) is in dollars per ton. Compute
producer’s surplus.
SOLUTION:
Since demand is 100 thousand tons, so
P(100) = 10+0.5(100)
= 10+50= 60 (thousands)
Therefore, producer’s surplus is
( )( )∫ +−
100
0
)5.01010060( dqqX
= 6000- [10q+0.5q2
/2] 0
100
= 6000-(1000+0.5X100X100/2)-0-0

10. =6000-(1000+2500)
= 6000-3500=2500 thousands
Graphical Presentation Of Producer / Suppliers surplus:
The area (yellow) above the Supply function and still in the rectangle
representing income is the producer surplus.

11. Consumer & producer surplus
Consumer
surplus
Producer
surplus
Equilibrium Price
Graphical Presentation Of Consumers and Producer / Suppliers surplus
Conclusion
From the above discussion we can say that, integral calculus is an
important part of business mathematics. We can use it in many ways and
take effective decisions. However it easily determines different terms
those can not be done easily in other ways. If anybody knows its
function and properties he/she can apply it to determine many important
things easily.