Introduction to Integral Calculus Along with Definition, Types, and Calculations

In mathematics, derivatives and integrals are two important concepts that are often used together. Integral calculus is a subject that many students choose to study in college, and for good reason.

It is an important tool that can be used to solve problems in various fields, including mathematics, physics, engineering, and biology. In this section, we will provide you with a brief introduction to integral calculus and its key concepts.

We will also discuss the process of integration and explain why it is necessary for students to learn this topic. Finally, we will provide you with a list of types of integrals and their corresponding rules

Integral Calculus: An Introduction

In the process of integration, integrals are found as values of the function. Integrating f(x) from f'(x) is called integration. In an integral, numbers are assigned to functions in a way that describes problems arising from displacement, motion, area, and volume. It is possible to determine the function f from the derivative f’ of the function f.

In order to understand integral calculus, students must first be familiar with basic concepts in mathematics. These concepts include variables, operations (such as addition and multiplication), and constants. In integrals, these same concepts are used to solve problems involving curves and surfaces. As we will see later, integration is essential for solving problems in many fields of study.

Integral calculus is a powerful technique that can be used to solve problems in many different fields of study. One example is engineering. Engineers use integrals to determine the behavior of objects under various conditions. This information is then used to create solutions for problems that may arise during the design process.

Integral calculus is also important for students studying biology. Biology relies heavily on mathematical models and equations to explain how biological systems work.

Integrals play an important role in these models by providing information about the dynamics of biological systems over time. Without a good understanding of integrals, biologists would not be able to make accurate predictions about how cells behave or how organisms interact with their surroundings.

Types of Integration

Integration can be defined in a number of ways, but for the purposes of this blog post we will focus on three main types:

  1. Limit integral
  2. Conditional integral
  3. Mutual Integral

Limit integration occurs when two systems are designed to work together in a specific way – for example, a system that controls the temperature in a factory and another system that monitors the output from the factory. In this type of integration, each system is designed to operate within its own boundaries.

Conditional integration allows one system to make decisions based on information obtained from another system – for example, if you have an online banking account with your bank, then your bank may be able to automatically transfer money into your account when you make a purchase. In this type of integration, both systems are reliant on each other – one cannot function without the other.

Mutual integration occurs when two or more systems share data – for example, all students at a school are enrolled in Google Maps so that they can see their location on maps. Mutual integration means that both systems benefit from sharing data – each system gets access to the data held by the other system and vice versa.

How to calculate the integral of a function?

For the numerical calculation of integral calculus, definite and indefinite integrals are used. These two topics are the methods of integral calculus for solving the numerical value of a function or a new function with or without using limits.

An integral calculator is an online source to solve the problems according to definite and indefinite integral.

Example 1: For indefinite integral

Evaluate the integral of the given function if the integrating variable is “w”.

f(w) = 4w3 + 3w5 – 4cos(w) – 4w + 5

Solution

Step 1: First of all, write the given function according to general integration form.

ʃ f(w) dw = ʃ [4w3 + 3w5 – 4cos(w) – 4w + 5] dw

Step 2: Now apply the sum and difference rules of integral calculus to write the notation of integration with each term of the given function.

ʃ [4w3 + 3w5 – 4cos(w) – 4w + 5] dw = ʃ [4w3] + ʃ [3w5] dw – ʃ [4cos(w)] dw – ʃ [4w] dw + ʃ [5] dw

Step 3: Now use the constant function rule of integral calculus.

ʃ [4w3 + 3w5 – 4cos(w) – 4w + 5] dw = 4ʃ [w3] + 3ʃ [w5] dw – 4ʃ [cos(w)] dw – 4ʃ [w] dw + ʃ [5] dw

Step 4: Now use the power and trigonometry rules of integral calculus to integrate the above expression.

ʃ [4w3 + 3w5 – 4cos(w) – 4w + 5] dw = 4 [w3+1 / 3 + 1] + 3 [w5+1 / 5 + 1] – 4 [sin(w)] – 4 [w1+1 / 1 + 1] + [5w] + C

= 4 [w4 / 4] + 3 [w6 / 6] – 4 [sin(w)] – 4 [w2 / 2] + [5w] + C

= 4/4 [w4] + 3/6 [w6] – 4 [sin(w)] – 4/2 [w2] + [5w] + C

= [w4] + 1/2 [w6] – 4 [sin(w)] – 2 [w2] + [5w] + C

= w4 + w6/2 – 4sin(w) – 2w2 + 5w + C

Example 2: For the definite integral

Evaluate the integral of the given function if the integrating variable is “t”.

f(t) = 4t + 9t2 – 2t3 – 11t + 2 in the interval of [0, 2]

Solution

Step 1: First of all, write the given function according to general integration form.

f(t) dt = [4t + 9t2 – 2t3 – 11t + 2] dt

Step 2: Now apply the sum and difference rules of integral calculus to write the notation of integration with each term of the given function.

[4t + 9t2 – 2t3 – 11t + 2] dt = [4t] dt + [9t2] dt – [2t3] dt – [11t] dt + [2] dt

Step 3: Now use the constant function rule of integral calculus.

[4t + 9t2 – 2t3 – 11t + 2] dt = 4[t] dt + 9[t2] dt – 2[t3] dt – 11[t] dt + [2] dt

Step 4: Now use the power and trigonometry rules of integral calculus to integrate the above expression.

[4t + 9t2 – 2t3 – 11t + 2] dt = 4 [t1+1/ 1 + 1]20 + 9 [t2+1/ 2 + 1]20 – 2 [t3+1/ 3 + 1]20 – 11 [t1+1/ 1 + 1]20 + [2t]20

= 4 [t2/ 2]20 + 9 [t3/ 3]20 – 2 [t4/ 4]20 – 11 [t2/ 2]20 + [2t]20

= 4/2 [t2]20 + 9/3 [t3]20 – 2/4 [t4]20 – 11/2 [t2]20 + 2 [t]20

= 2 [t2]20 + 3 [t3]20 – 1/2 [t4]20 – 11/2 [t2]20 + 2 [t]20

Step 5: Now use the fundamental theorem of calculus to apply the upper and lower limit values to the above-integrated value.

= 2 [22 – 02] + 3 [23 – 03] – 1/2 [24 – 04] – 11/2 [22 – 02] + 2 [2 – 0]

= 2 [4 – 0] + 3 [8 – 0] – 1/2 [16 – 0] – 11/2 [4 – 0] + 2 [2 – 0]

= 2 [4] + 3 [8] – 1/2 [16] – 11/2 [4] + 2 [2]

= 8 + 24 – 16/2 – 44/2 + 4

= 8 + 24 – 8 – 44/2 + 4

= 8 + 24 – 8 – 22 + 4

= 32 – 8 – 22 + 4

= 24 – 22 + 4

= 2 + 4

= 6

Wrap Up

Integral is the fundamentals of calculus that are used to evaluate the area under the curve. This topic is very essential for students in grades 11 and 12. The students who are not familiar with this concept can take assistance from this post.

   

Leave a Reply

Your email address will not be published. Required fields are marked *

Thank you for your interest
Please leave your details to get the best colleges and free counseling
Thank you for your interest
Please leave your details to get the best colleges and free counseling