Introduction to the types of integrals explained with examples

Introduction to the types of integrals explained with examples

In calculus, the types of integral are frequently used to solve the problems of integration. Integral is a branch of calculus used to find the area under the curve and the volume. Limits of calculus are widely used to define the integral, especially the definite integral.

In this post, we will learn the definition, types, and formulas of integration along with a lot of examples.

What are the types of integral?

Before discussing the types of integrals, you have a sound knowledge of integrals.

In calculus, the integral is used to find the total size or values such as volumes, lengths, and areas. This type of calculus is connected with the fundamental theorem of calculus, which shows how to evaluate the definite integral.

It is the inverse process of derivatives and is used to integrate the functions. There are two types of integration in calculus:

  • Definite integral
  • Indefinite integral

Let’s discuss the types of integral briefly.

(i)          The definite integral

The definite integral is a type of integral used to find the numerical values of the given functions by using the fundamental theorem of calculus. In this type of integration, the interval values of the functions exist such as [a, b].

The first term of the interval will be taken as a lower limit of the function and the next term will be the upper limit. The upper and lower limits must be written along with the integral notation. The general formula of the definite integral is:

f(z) dz = F(m) – F(n) = K

  • In the above equation, m & n are the upper and lower limits of the function.
  • f(z) is the function to be integrated.
  • z is the corresponding variable or integrating variable.
  • F(m) – F(n) is the fundamental theorem of calculus used to apply the upper and lower limits of the function.
  • K will be the numerical result after applying the fundamental theorem of calculus.

(ii)        The indefinite integral

This is the other type of integral which is used to integrate the given function without using the upper and the lower limits. By using the rules of integration, the definite integral solves the problems of integral.

The formula of the indefinite integral is:

ʃ f(z) dz = F(z) + C

  • f(z) is the function to be integrate.
  • z is the corresponding variable or integrating variable.
  • F(z) is the result after integrating the function.
  • C is the constant of integration.

Examples of Definate and indefinite integrals

Below are solved examples of the definite and indefinite integral.

Example 1: For indefinite integral

Integrate 12z3 + 14sin(z) – 24x2z + 16y + 4 with respect to z.

Solution

Step 1: Write the given function along with the integral notation of indefinite integral.

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz

Step 2: Apply the indefinite integral notation separately to each function

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz = ʃ (12z3) dz + ʃ 14sin(z) dz – ʃ 24x2z dz + ʃ 16y dz + ʃ 4 dz

Step 3: Write the constant coefficient of the given expression outside the integral notation by using the constant function rule.

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz =12 ʃ (z3) dz + 14 ʃ (sin(z)) dz – 24x2 ʃ (z) dz + 16y ʃ dz + 4 ʃdz

Step 4: Integrate the above expression by using the power and trigonometric rules of integration.

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz =12 (z3+1 / 3 + 1) + 14 (-cos(z)) – 24x2 (z1+1/ 1+ 1) + 16y (z) + 4 (z) + C

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz =12 (z4 / 4) + 14 (-cos(z)) – 24x2 (z2/ 2) + 16y (z) + 4 (z) + C

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz =12/4 (z4) + 14 (-cos(z)) – 24x2/2 (z2) + 16y (z) + 4 (z) + C

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz =3 (z4) – 14cos(z) – 12x2 (z2) + 16y (z) + 4 (z) + C

ʃ (12z3 + 14sin(z) – 24x2z + 16y + 4) dz =3z4 – 14cos(z) – 12x2z2 + 16yz + 4z + C

In the above example we have calculate the indefiniate integrals by applying the rules of integration on the function.

The above problem can also be solved by a integral calculator which can ease up the process of evaluating integrals and help you to understand and get the solution of indefinite integrals with step by step explanation in a couple of second

Step 1: Select the type of integral.

Step 2: Input the function.

Step 3: Choose the integrating variable.

Step 4: Click the calculate button.

Step 5: The result with steps will show below the calculate button.

Example 2: For definite integral

Integrate 5z3 + 2sin(z) – 15y4z3 + 5z2 + 2 with respect to x and the boundaries are from 3 to 4.

Solution

Step 1: Write the given function along with the integral notation of indefinite integral.

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz

Step 2: Apply the indefinite integral notation separately to each function

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = (5z3) dz +2sin(z) dz – 15y4z3 dz + 5z2 dz + 2 dz

Step 3: Write the constant coefficient of the given expression outside the integral notation by using the constant function rule.

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 5 (z3) dz + 2sin(z) dz – 15y4 z3 dz + 5 z2 dz + 2  dz

Step 4: Integrate the above expression by using the power and trigonometric rules of integration.

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 5 [(z3+1 / 3 + 1)]43 + 2 [-cos(z)]43 – 15y4 [z3+1 / 3 + 1]43 + 5 [z2+1 / 2 + 1]43 + 2 [z]43

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 5 [(z4/ 4)]43 + 2 [-cos(z)]43 – 15y4 [z4 / 4]43 + 5 [z3 / 3]43 + 2 [z]43

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 5/4 [z4]43 – 2 [cos(z)]43 – 15y4/4 [z4]43 + 5 /3[z3]43 + 2 [z]43

Step 5: Use f(z) dz = F(m) – F(n) to find the definite integral.

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 5/4 [44 – 34] – 2 [cos(4) – cos(3)] – 15y4/4 [44 – 34] + 5 /3 [43 – 33] + 2 [4 – 3]

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 5/4 [256 – 81] – 2 [cos (4) – cos (3)] – 15y4/4 [256 – 81] + 5 /3 [64 – 27] + 2 [4 – 3]

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 5/4 [175] – 2 [cos (4) – cos (3)] – 15y4/4 [175] + 5 /3 [37] + 2 [1]

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 875/4 – 2cos (4) + 2cos (3) – 2625y4/4 + 185/3 + 2

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 218.75 – 2cos (4) + 2cos (3) – 656.25y4 + 61.67 + 2

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 218.75 – 2cos (4) + 2cos (3) – 656.25y4 + 63.67

(5z3 + 2sin(z) – 15y4z3 + 5z2 + 2) dz = 282.42 – 2cos (4) + 2cos (3) – 656.25y4

Summary

In this article, we have learned the types of integral with examples. After reading the above article now you are able to solve any type of integral whether it is definite or indefinite integral by applying the rules of integration or solving and understanding it using antiderivative calculator.

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