Mathematics is a wide concept in the sciences. There are various branches of mathematics such as calculus, algebra, geometry, number theory, trigonometry, etc. All these branches play a vital role in educational and daily life calculations.
In this post, we are going to describe calculus along with its types and its calculations.
Calculus – Definition
In mathematics, a term that deals with the properties and evaluation of differential & integral of single, double, or multivariable functions, by methods originally based on the summation of infinitesimal differences.
Kinds of calculus
There are three main and wellknown kinds of calculus such as:
 Limit calculus
 Integral calculus
 Differential calculus
Let us discuss the kinds of calculus along with definitions, formulas, rules, & examples.

Limit calculus
In calculus, the Limit is widely used to define the other branches of calculus such as continuity, integral, and differential. The limit calculus is defined as a value of a function that approaches some value by placing the specific point in the function.
Formula of Limit Calculus
The general expression of limit calculus is:
Limx→a k(x) = N

 “x” = independent variable of the function.
 k(x) = given function
 “a” = the limit value that a function approaches.
 N = final result after placing the particular value.
Kinds of limit calculus
 Lefthand limit (LHL)
 Righthand limit (RHL)
 Twosided limit (TSL)
Example of Limit Calculus
Example
Calculate the limit of the given function if the independent variable x approaches 2.
k(x) = 2.5x3 – 5.5x4 + 12x +150
Solution
Step 1: Firstly, take the given function and apply the notation of the limit on it.
k(x) = 2.5x3 – 5.5x4 + 12x +150
limit value = a = 2
limx→a k(x) = limx→2 [2.5x3 – 5.5x4 + 12x +150]
Step 2: Now apply the notation of limit calculus to each function separately with the help of the sum and difference rules of limits.
limx→2 [2.5x3 – 5.5x4 + 12x +150] = limx→2 [2.5x3] – limx→2 [5.5x4] + limx→2 [12x] + limx→2 [150]
Step 3: Now take the constant coefficient outside the limit notation.
limx→2 [2.5x3 – 5.5x4 + 12x +150] = 2.5 limx→2 [x3] – 5.5 limx→2 [x4] + 12 limx→2 [x] + limx→2 [150]
Step 4: Now apply the power and constant rules of limits.
= 2.5 [23] – 5.5 [24] + 12 [2] + [150]
= 2.5 [8] – 5.5 [16] + 12 [2] + [150]
= 20 – 88 + 24 + 150
= – 68 + 24 + 150
= – 44 + 150
= 106
You can cross check the result with the help of a limit solver.

Integral calculus
In calculus, the integral is used to determine the new function or a numerical value of the function. It is usually used to determine the area under the curve. The upper and lower limit values are involved in this type of calculus.
Kinds of integral calculus
 Definite integral
 Indefinite integral
Formula of Integral Calculus
The formula for the definite integral is:
ab k(x) dx = K(b) – K(a) = M
The formula for the indefinite integral is:
ʃ k(x) dx = K(x) + C
 ʃ = integral notation
 a & b = boundary values
 k(x) is the integrand function.
 “x” = the integrating variable.
 K(b) – K(a) = applying limit values by the fundamental theorem of calculus.
 M = final result of definite integral
 C = constant of integration
An integral calculator with steps is a helpful tool to integrate the definite and indefinite integrals according to the above formulas with steps.
Example of Integral Calculus
Example
Find the integral of the given function with respect to “x”.
k(x) = 3.5x4 + 4.5x – 2.5x4 + 120
Solution
Step 1: First of all, apply the integral notation to the given function.
k(x) = 3.5x4 + 4.5x – 2.5x4 + 120
ʃ k(x) dx = ʃ [3.5x4 + 4.5x – 2.5x4 + 120] dx
Step 2: Use the sum and difference rule of integral calculus and apply the integral calculus notation to each function.
ʃ [3.5x4 + 4.5x – 2.5x4 + 120] dx = ʃ [3.5x4] dx + ʃ [4.5x] dx – ʃ [2.5x4] dx + ʃ [120] dx
Step 3: Use the constant function rule of integral and take the constant coefficients outside the notation.
ʃ [3.5x4 + 4.5x – 2.5x4 + 120] dx = 3.5ʃ [x4] dx + 4.5ʃ [x] dx – 2.5ʃ [x4] dx + ʃ [120] dx
Step 4: Integrate the above expression.
= 3.5 [x4+1 / 4 + 1] + 4.5 [x1+1 / 1 + 1] – 2.5 [x4+1 / 4 + 1] + [120x] + C
= 3.5 [x5 / 5] + 4.5 [x2 / 2] – 2.5 [x5 / 5] + [120x] + C
= 3.5/5 [x5] + 4.5/2 [x2] – 2.5/5 [x5] + [120x] + C
= 35/50 [x5] + 45/20 [x2] – 25/50 [x5] + [120x] + C
= 7/10 [x5] + 9/4 [x2] – 1/2 [x5] + [120x] + C
= 2x5/10 + 9x2/4 + 120x + C
= x5/5 + 9x2/4 + 120x + C
Derivative Calculus
In calculus, the derivative is widely used to calculate the slope of the tangent line. The derivative calculus is also defined as the rate of change of single, double, or multivariable functions with respect to the independent variable.
Formula of Derivative
f’(x) = Limh→0 [f(x + h) – f(x)]/h
Kinds of Derivative
 Explicit derivatives
 Directional derivative
 Implicit differentiation
 Partial derivative
Example of Derivative
Example
Calculate the derivative of the given function with respect to “y”.
K(y) = 5.5y3 + 2.5y5 – 8.5y4 + 3y + 900
Solution
Step I: First of all, write the given differential function and apply the notation of differentiation to it.
K(y) = 5.5y3 + 2.5y5 – 8.5y4 + 3y + 900
d/dy [k(y)] = d/dy [5.5y3 + 2.5y5 – 8.5y4 + 3y + 900]
Step II: Now apply the notation of differentiation to each function separately by using the sum and difference rules of differentiation.
d/dy [5.5y3 + 2.5y5 – 8.5y4 + 3y + 900] = d/dy [5.5y3] + d/dy [ 2.5y5] – d/dy [8.5y4] + d/dy [3y] + d/dy [900]
Step III: Now apply the constant function rule of differential calculus.
d/dy [5.5y3 + 2.5y5 – 8.5y4 + 3y + 900] = 5.5d/dy [y3] + 2.5d/dy [y5] – 8.5d/dy [y4] + 3d/dy [y] + d/dy [900]
Step IV: Now find the derivative of the above expression with respect to “u”.
= 5.5 [3y31] + 2.5 [5y51] – 8.5 [4y41] + 3 [y11] + [0]
= 5.5 [3y2] + 2.5 [5y4] – 8.5 [4y3] + 3 [y0] + [0]
= 5.5 [3y2] + 2.5 [5y4] – 8.5 [4y3] + 3 [1]
= 16.5 [y2] + 12.5 [y4] – 34.0 [y3] + 3 [1]
= 16.5y2 + 12.5y4 – 34y3 + 3
Wrap up
Now by learning this post, you are able to solve calculus problems easily. The main types of calculus (limit, derivative, & integral) along with explanations and calculations are described in the above post.