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Introduction to Calculus: Definition, Types, and Examples

Introduction to Calculus: Calculus Definition, Calculus Types, Calculus Examples and Calculus Formulas, Introduction to Calculus PDF, Calculus Notes PDF.

Calculus is one of the main branches of mathematics. The other branches are algebra, geometry, trigonometry, etc. mathematics is a wide concept in sciences used to solve numerical problems and daily life calculations.

In this post, we will discuss the main branches of calculus along with definitions, types, and examples.

What is calculus?

Calculus is a branch of mathematics that deals with the properties and evaluation of integration and differentiation of functions, by methods originally based on the summation of infinitesimal differences.

Types of calculus

There are various types of calculus such as:

  • Limit calculus
  • Integral calculus 
  • Differential calculus

These types are frequently used as a system of calculation or reasoning. Let us briefly discuss the types of calculus with examples.

  • Limit calculus

Limit is a branch of calculus that is widely used to define the other branches of calculus such as integration and differential. It is defined as a value that tells a function approach to some value by placing the particular value in the function. 

Limit formula

The formula of the limit calculus is:

limh→n p(h) = M

    •  “h” = a variable of the function.
    • P(h) = given function
    • “n” = the limit value that a function approaches.
  • M = final result after placing the particular value. 

A limit calculator with steps can be used to get the step-by-step solution to limit calculus problems according to the above formula.

Types of limits

  1. Left-hand limit
  2. Right-hand limit
  3. Two-sided limit

Example of limit

Example

Calculate the limit function p(h) = 2h2 – 5h3 + 3h +15, when h approaches 3.

Solution 

Step 1: First of all, take the given function and apply the notation of the limit on it.

p(h) = 2h2 – 5h3 + 3h +15

limit value = n = 3

limh→3 p(h) = limh→3 [2h2 – 5h3 + 3h +15]

Step 2: Now apply the notation of limit calculus to each function separately with the help of the sum and difference rules of limits. 

limh→3 [2h2 – 5h3 + 3h +15] = limh→3 [2h2] – limh→3 [5h3] + limh→3 [3h] + limh→3 [15]

Step 3: Now take the constant coefficient outside the limit notation. 

= 2limh→3 [h2] – 5limh→3 [h3] + 3limh→3 [h] + limh→3 [15]

Step 4: Now apply the power and constant rules of limits. 

= 2 [32] – 5 [33] + 3 [3] + [15]

= 2 [9] – 5 [270] + 3 [3] + [15]

= 18 – 135 + 9 + 15

= -117 + 9 + 15

= -93

  • Integral calculus 

The integral calculus is frequently used to find the area under the curve. It is defined as a new function whose original function is a differential or a numerical value by taking the boundary values.

Integral formula

The formulas for the integral are:

  • ab p(h) dh = P(b) – P(a) = N
  • ʃ p(h) dh = P(h) + C

where,

  • ʃ = integral notation 
  • a & b = boundary values
  • p(h) is the integrand function.
  • “h” = the integrating variable.
  • P(v) – P(u) = applying limit values by the fundamental theorem of calculus. 
  • N = final result
  • C = constant of integration

Types of integral calculus

  • Definite integral
  • Indefinite integral 

Example of integral 

Example 

Find the integral of p(h) = 3h3 + 4h – 2h4 + 1 with respect to h. 

Solution 

Step 1: First of all, apply the integral notation to the given function.

p(h) = 3h3 + 4h – 2h4 + 1 

ʃ p(h) dh = ʃ [3h3 + 4h – 2h4 + 1] dh 

Step 2: Use the sum and difference rule of integral calculus and apply the integral calculus notation to each function.

ʃ [3h3 + 4h – 2h4 + 1] dh = ʃ [3h3] dh + ʃ [4h] dh – ʃ [2h4] dh + ʃ [1] dh

Step 3: Use the constant function rule of integral and take the constant coefficients outside the notation.

= 3ʃ [h3] dh + 4ʃ [h] dh – 2ʃ [h4] dh + ʃ [1] dh 

Step 4: Integrate the above expression. 

 = 3 [h3+1 / 3 + 1] + 4 [h1+1 / 1 + 1] – 2 [h4+1 / 4 + 1] + [h] + C 

 = 3 [h4 / 4] + 4 [h2 / 2] – 2 [h5 / 5] + [h] + C 

 = 3/4 [h4] + 4/2 [h2] – 2/5 [h5] + [h] + C 

 = 3/4 [h4] + 2 [h2] – 2/5 [h5] + [h] + C 

 = 3h4/4 + 2h2 – 2h5/5 + h + C 

To avoid these calculations to find the integral of the function use an antiderivative calculator.

  • Differential Calculus 

The differential calculus is frequently used to find the slope of the tangent line. It is defined as the instantaneous rate of change of a function with respect to the independent variable of the function. It can also be defined by using the limits.

It is denoted by d/dx.

Derivative formula

f’(x) = Limh→0 [f(x + h) – f(x)]/h

Types of differential calculus

  • Explicit derivatives
  • Implicit differentiation  
  • Partial derivative

Example of differential calculus 

Example 

Calculate 5h2 + 2h3 – 8h4 + 3h + 9 with respect to “h”.

Solution 

Step I: First of all, write the given differential function and apply the notation of differentiation to it. 

p(h) = 5h2 + 2h3 – 8h4 + 3h + 9

d/dh [p(h)] = d/dh [5h2 + 2h3 – 8h4 + 3h + 9]

Step II: Now apply the notation of differentiation to each function separately by using the sum and difference rules of differentiation. 

d/dh [p(h)] = d/dh [5h2 + 2h3 – 8h4 + 3h + 9] = d/dh [5h2] + d/dh [2h3] – d/dh [8h4] + d/dh [3h] + d/dh [9]

Step III: Now apply the constant function rule of differential calculus. 

= 5 d/dh [h2] + 2 d/dh [h3] – 8 d/dh [h4] + 3 d/dh [h] + d/dh [9]

Step IV: Now find the derivative of the above expression with respect to “u”.

= 5 [2 * h2-1] + 2 [3 * h3-1] – 8 [4 * h4-1] + 3 [h1-1] + [0]

= 5 [2 * h1] + 2 [3 * h2] – 8 [4 * h3] + 3 [h0] + [0]

= 10 * h + 6 * h2 – 32 * h3 + 3 [1] + [0]

= 10 * h + 6h2 – 32h3 + 3

Conclusion 

In this post, we have covered some well-known types of calculus along with definitions, types, and solved examples. Now you can solve any problem of differential, integral, and limit just by following the above basics of these branches.

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