Integral calculus is the branch of mathematics that deals with integrals and their characteristics. Moreover, it provides interesting information about the antiderivatives of functions.
It also provides the information determination of integral properties, and their applications. Integration is a concept that finds the area under the curve. For discussion of integral we have detailed knowledge of derivative or differentiation.
In this article, we will thoroughly discuss integral calculus, along with its definition, types, formulas, rules, and for a better understanding of integral solving different examples.
Definition of Integral:
“Integral is the continuous analog of a sum, which is used to calculate Areas, volumes and their generalizations.” It’s represented as “∫”.
Integral Calculus:
If we know the g’ of the functions that are differentiable in their domains, then we can calculate g. In Differentiate calculus g’ is recalling the derivative of the function g. On the other hand, G is related to the antiderivative of g’ in integral calculus.
Antidifferentiation or integration are terms used to describe the process of identifying the antiderivatives. It is the reverse process of derivatives.
Types of integral:
Generally, there are two basic types of integral.
 Indefinite Integral
 Definite Integral
Definite Integral:
In a definite integral, the lower and upper bound of the integral is defined i.e. (start and end value). It is defined as,
∫ba f(x) dx = F(b) – F(a) = L
Where L is a constant number solution of the definite integral.
Indefinite Integral:
In this integral, the lower and upper bounds are not defined. The integration of f(x) is F(x), and its formula can be represented as
∫ f(x) dx = F(x) + C Where C is an integrating constant that comes in indefinite integral.
Rules of Integral:
There are some basic rules of integral discussed below in the table:
Rules  General Formula 
Sum Rule  ∫ [f(x) + g (x)] dx = ∫ [f(x)] dx+ ∫ [g(x)] dx 
Difference Rule  ∫ [f(x) – g (x)] dx = ∫ [f(x)] dx – ∫ [g(x)] dx 
Power Rule

∫ [f(x)] n dx = [f(x)] n+1 / (n + 1) + c 
Constant function Rule  ∫ [Q f(x)] dx = Q ∫ [f(x)] dx 
Formulas
Some useful trigonometric and exponential formulas are described in the following table.
Functions  Formula 
Cos(x)  ∫ [cos(x)] dx = sin (x) + c 
Sin (x)  ∫ [sin(x)] dx = sin (x) + c 
Tan(x)  ∫ [tan(x)] dx = ln(cos(x)) + c 
Sin1(x)  ∫ dx/ (b2 – x2) ½ = [ sin 1(x/b)] + c 
Cos1(x)  ∫dx/ (b2 – x2)1/2 = [ cos1(x/b)] + c 
Tan1(x)  ∫ dx/ (b2 + x2) = 1/b [ tan1(x/b)] + c 
Exponential  ∫ enx dx = enx / n + c
Where “c” is a constant. 
Calculations of Integral Calculus
The calculations of the integral calculus problems can be done either with the help of rules and formulas by solving them manually or with the help of integration calculator to get the result in a few seconds. Here are a few examples to understand the calculations of integral calculus manually.
Example 1:
Find the integral of z2 (z – 1).
Solution:
Step 1: Let the given value is equal to A(z).
A(z) = z2 (z – 1)
A(z) = z3 – z
Step 2: Apply integral on both sides w.r.t the “z”
∫ A(z) dz = ∫ (z3 – z) dz
Step 3: Apply the Difference rule and separate the integral.
∫ [f(x) – g (x)] dx = ∫ [f(x)] dx – ∫ [g(x)] dx (Difference rule)
∫ A(z) dz = ∫ z3 dz – ∫ z dz
Step 4: solve the integral by power rule.
∫ [G(z)] n dz = [G(z)] n+1 / (n + 1) + C
∫ A(z) dz = z4 / 4 – z2/2 + C
Thus, ∫ z2 (z – 1) dz = (z4 / 4) – (z2/2) + C
Example 2:
Find the integral of 3 Tan(x) + 4 cos(x).
Solution:
Step 1: let the given value be equal to a function.
B(x) = 3 Tan(x) + 4 cos(x)
Step 2: Apply the integral on both sides.
∫ B(x) dx = ∫ [3 Tan(x) + 4Cos(x)]
Step 3: separate the integral with the help of the sum rule.
∫ [f(x) + g (x)] dx = ∫ [f(x)] dx+ ∫ [g(x)] dx
∫ B(x) dx = ∫ [ 3 Tan(x)] dx + ∫ [4 cos(x)] dx
Step 4: Take out the Constant from the integral by constant function rule.
∫ B(x) dx =3 ∫ [Tan(x)] dx + 4 ∫ [ cos(x)] dx
Step 5: Apply the integral formula of “Tan(x)” & “cos(x)”.
∫ B(x) dx = 3[ ln(cos(x))] + 4 [ sin (x)] + c
∫ B(x) dx = – 3 ln(cos(x)) + 4 [ sin (x)] + c
∫ 3 Tan(x) + 4 cos(x) dx = 3 ln(cos(x)) + 4 [sin (x)] + c
Example 3:
Find the solution of (e2x + 2x) dx.
Solution:
Step 1: Let the given integral is equal to F(x).
E(x) = (e2x + 2x) dx
Step 2: Separate the integral by the sum rule.
∫ [h(x) + I(x)] dx = ∫ [h(x)] dx + ∫ [I(x)] dx
E(x) = (e2x) dx + (2x) dx
Step 3: Use the integral formula of an exponential and algebraic function.
∫ enx dx = enx / n
E(x) = (1/2) e2x 02 + 2(x2/2) 02
E(x) = (1/2) e2x 02 + x2 02
Step 4: Put the value of upper & lower limits and simplify.
E(x) = 1/2[e0 – e4] + [(0)2 – (2)2]
E(x) = 1/2[1 e4] + [–4]
E(x) = 1/2[1 e4] – 4
E(x) = 1/2 (1/2) e4 – 4
E(x) = [1/2 – 4] – (1/2) e4
E(x) = [1/2 – 4] – (1/2) e4
E(x) = [1– 8/2] – (1/2) e4
E(x) = [– 7/2] – (1/2) e4
E(x) = 1/2 [7+ e4]
(e2x + 2x) dx = 1/2[7 + e4].
Summary:
In this article, we discussed the basic definition of integral calculus and its types. Further, we have discussed some basic formulas and rules of integration. In the example section, we solved the definite and indefinite integral by using the rules of integration.
Integral calculus is a very interesting topic, hope you have gone through the basic concept of this topic by reading this article and easily solved the related problems.