Business Application of Integral Calculus
Dr. Rashed Al Karim
PhD (UUM); MBA (UK);
M.Sc. (CU); B.Sc. (CU)
Page | 1
Definition: If φ(x) be any differentiable function of ‘x’ such that
𝑑
𝑑𝑥 [φ(x)] = f(x);
then φ(x) is called an anti-derivative or an integral of f(x). Symbolically, it can be written as:
∫𝑓(𝑥)𝑑𝑥 = φ(x)
and is read as “φ(x) is the integral of f(x) with respect to x”.
Thus, Integral calculus deals with the inverse operation of differentiation i.e., anti-derivative.
The process of finding the integral of a given function is called integration and the given
function is called the integrand.
There are mainly two types of Integration such as: Indefinite Integral and Definite Integral.
Indefinite Integral: This type of integration does not give a definite value. For example,
∫2𝑥𝑑𝑥 = x2+ c; where ‘c’ is an arbitrary constant, called the constant of integration.
Definite Integral: This type of integration gives a definite value and can be written as
∫𝑓(𝑥)𝑑𝑥
𝑏
𝑎 = [φ(𝑥)] 𝑎
𝑏 = φ(𝑏) – φ(𝑎)
Where, a and b are two real numbers, and called the lower limit and the upper limit of the
integral respectively.
Some Standard Formula of Integral Calculus:
(i) ∫𝑑𝑥 = x (ii) ∫𝑥𝑛𝑑𝑥 = 𝑥𝑛+1
𝑛+1
(iii) ∫1
𝑥 𝑑𝑥 = logx (iv) ∫𝑒𝑥𝑑𝑥 = 𝑒𝑥
(v) ∫𝑚𝑒𝑚𝑥 𝑑𝑥 = 𝑒𝑚𝑥 (vi)∫𝐶𝑜𝑠𝑥 𝑑𝑥 = Sinx
(vii) ∫𝑆𝑖𝑛𝑥 𝑑𝑥 = – Cosx
Example-1: Integrate the following with respect to ‘x’
(i) ∫(3 − 2𝑥 − 𝑥4)𝑑𝑥 (ii) ∫(𝑥2− 1)2𝑑𝑥
Solution: (i) ∫(3 − 2𝑥 − 𝑥4)𝑑𝑥 = ∫3𝑑𝑥 – ∫2𝑥𝑑𝑥 – ∫𝑥4𝑑𝑥
= 3∫𝑑𝑥 – 2∫𝑥𝑑𝑥 – ∫𝑥4𝑑𝑥
= 3x – 2. 𝑥1+1
1+1 – 𝑥4+1
4+1
Business Application of Integral Calculus
Dr. Rashed Al Karim
PhD (UUM); MBA (UK);
M.Sc. (CU); B.Sc. (CU)
Page | 2
= 3x – x2 – 𝑥5
5 [Ans]
(ii) ∫(𝑥2− 1)2𝑑𝑥 = ∫(𝑥4− 2. 𝑥2. 1 + 1) 𝑑𝑥
=∫(𝑥4− 2𝑥2+ 1) 𝑑𝑥
=∫𝑥4 𝑑𝑥 – 2∫𝑥2𝑑𝑥 +∫𝑑𝑥
= 𝑥 4+1
4+1 – 2 𝑥 2+1
2+1 + x
= 𝑥5
5 – 2
3 𝑥3 + x [Ans]
Example-2: Integrate the following with respect to ‘x’
(i) ∫(4𝑥2
2
0− 8𝑥)𝑑𝑥 (ii) ∫(3𝑥 − 2)2𝑑𝑥
4
2
Solution: (i) ∫(4𝑥2
2
0− 8𝑥)𝑑𝑥 = ∫4𝑥2
2
0𝑑𝑥 –∫8𝑥
2
0𝑑𝑥
= 4[ 𝑥2+1
2+1 ]0
2 – 8[ 𝑥1+1
1+1 ]0
2
= 4
3[ 𝑥3]0
2 – 4 [ 𝑥2]0
2
= 4
3[2 3 – 0] – 4 [ 2 2– 0]
= 4
3 .8 – 4.4 = 32
3 – 16 = 32 − 48
3 = − 16
3 [Ans]
(ii) ∫(3𝑥 − 2)2𝑑𝑥
4
2= ∫(9𝑥2− 2.3𝑥. 2 + 4)𝑑𝑥
4
2
= ∫(9𝑥2𝑑𝑥 − 12𝑥 𝑑𝑥 + 4𝑑𝑥)
4
2
= [9. 𝑥3
3 – 12. 𝑥2
2 +4x ]2
4
= [3𝑥3– 6𝑥2 +4x ]2
4
= [3.43 – 6.42 + 4.4] – [3.23 – 6.22 + 4.2]
= [3.64 – 6.16 +16] – [3.8 – 6.4 + 8]
= (192 – 96+16) – (24 – 24 + 8)
= 112 – 8 = 104 [Ans]
Business Application of Integral Calculus
Dr. Rashed Al Karim
PhD (UUM); MBA (UK);
M.Sc. (CU); B.Sc. (CU)
Business Application
We know that if the total cost function (TC) is given then the marginal cost function (MC) is
the first derivative of the total cost function (TC). It follows, therefore, that the total cost
function (TC) is the integral of the marginal cost function (MC).
Thus, if TC represent the total cost of producing an output ‘x’ then the marginal cost function
is given by MC = 𝑑𝑇𝐶
𝑑𝑥
∴
Total Cost, TC =∫(𝑴𝑪)𝒅𝒙 + 𝑲; where, k is the constant of integration, can be
evaluated if the fixed cost (i.e., the cost when x = 0) is given.
Further, the average cost (AC) can be obtained from the relation: AC = 𝑻𝑪
𝒙 .
Similarly, Marginal Revenue, MR = 𝑑𝑇𝑅
𝑑𝑥 ;
∴
Total Revenue, TR = ∫(𝑴𝑹)𝒅𝒙 + 𝑲
Again, we know, Total Revenue, TR = unit price × output = p x
