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In mathematics, the integral is one of the essential types of calculus that is frequently used to find the area under the curve and antiderivative of the function. This branch of calculus is frequently used to find the volume and other quantities that are difficult to evaluate by other methods.

It is an essential type of calculus along with differential calculus that deals with the instantaneous rate of change. In this article, we will discuss the term integral calculus along with its properties and types with solved examples.

## What is integral in calculus?

In mathematics, calculus is frequently used for dealing with change and motions. There are two well-known branches of calculus such as integral calculus and differential calculus. The integral calculus is used to find the antiderivative, area under the curve, and volume.

While the other type of calculus is used to calculate the rate of change of the function with respect to the independent variable. The integral is the most frequently used type of calculus that is essential for the calculations of various things to get rid of difficult ways.

## Types of Integral Calculus

There are two types of integral calculus such as antiderivative and definite integral. Here is a brief introduction to the types of the integral calculus.

### Antiderivative

The antiderivative is the branch of integral calculus that is essential to find the new function whose original function is a differential function with respect to integrating variables. The constant of integration will be applied with the integral of the function.

There are no boundary values involved in this type of integral calculus. The term antiderivative is also known as indefinite integral. The expression of antiderivative is:

∫ f(t) dt = F(t) + C

Where,

·         f(t) = differential function

·         F(t) = new function

·         t = integrating variable

·         C = integrating constant

### Definite integral

The other type of integral calculus is definite integral which is used to find the area and volume under the curve with the help of boundary values of the function. The boundary values of the function are known as the upper and lower limit values.

The boundary values are to be applied to the new function with the help of the fundamental theorem of calculus. The expression of the definite integral is:

uv f(t) dt = [F(t)]vu = F(v) – F(u)

Where,

·         u & v = upper and lower limit values

·         f(t) = differential function

·         F(t) = new function

·         t = integrating variable

## Properties of Integral Calculus

There are several properties of integral calculus. Here we are going to discuss some of them.

 Properties Expressions Power Property ʃ f(t) n dt = f(t) n + 1/ (n + 1) + C Constant Property ʃ K dt = K * t + C Constant Function Property ʃ K * f(t) dt = K * ʃ f(t) dt + C Sum Property ʃ [f(t) + g(t)] dt = ʃ [f(t)] dt + ʃ [g(t)] dt + C Difference Property ʃ [f(t) - g(t)] dt = ʃ [f(t)] dt - ʃ [g(t)] dt + C Trigonometry Property ʃ sin(t) dt = -cos(t) + C ʃ cos(t) dt = sin(t) + C ʃ tan(t) dt = ln |sec(t)| + C Exponential Property ʃ et dt = et + C Product Property ʃ f(t) * g’(t) dt = f(t) g(t) - ʃ g(t) * f’(t) dt

## How to calculate integral calculus problems?

There are two ways of calculating integral calculus problems such as solving them with the help of an online integral calculator and the other is solving them manually with the help of properties and formulas.

Example 1

Calculate the antiderivative of g(t) = 12t2 + 2cos(t) – 4t3 + 12t – 100 with respect to “t”

Solution

Using manual method

Step 1: Write the given differential function and apply the notation of antiderivative to it.

ʃ g(t) dt = ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt

Step 2: Now apply the sum and difference properties of integral calculus to the above expression.

ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt = ʃ [12t2] dt + ʃ [2cos(t)] dt – ʃ [4t3] dt + ʃ [12t] dt – ʃ  dt

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

ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt = 12ʃ [t2] dt + 2ʃ [cos(t)] dt – 4ʃ [t3] dt + 12ʃ [t] dt – ʃ  dt

Step 4: Now integrate the above expression with the help of power and trigonometry properties.

ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt = 12 [t2 + 1 / 2 + 1] + 2 [sin(t)] – 4 [t3 + 1 / 3 + 1] + 12 [t1 + 1 / 1 + 1] – [100 t] + C

ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt = 12 [t3 / 3] + 2 [sin(t)] – 4 [t4 / 4] + 12 [t2 / 2] – 100 [t] + C

ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt = 12/3 [t3] + 2 [sin(t)] – 4/4 [t4] + 12/2 [t2] – 100 [t] + C

ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt = 4 [t3] + 2 [sin(t)] – 1 [t4] + 6 [t2] – 100 [t] + C

ʃ [12t2 + 2cos(t) – 4t3 + 12t – 100] dt = 4t3 + 2sin(t) – t4 + 6t2 – 100t + C

Using calculator the result will be:

Example 2

Calculate the area under the curve between the interval of [2, 3] with respect to “v”

Example 1

Calculate the numerical value of g(t) = 2t3 + 6t2 – 12t5 + 16t – 4t4 with respect to “t” in the interval of [2, 3]

Solution

Step 1: Write the given differential function and apply the notation of integral to it with the interval.

uʃv g(t) dt = 2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt

Step 2: Now apply the sum and difference properties of integral calculus to the above expression.

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 2ʃ3 [2t3] dt + 2ʃ3 [6t2] dt – 2ʃ3 [12t5] dt + 2ʃ3 [16t] dt – 2ʃ3 [4t4] dt

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

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 22ʃ3 [t3] dt + 62ʃ3 [t2] dt – 122ʃ3 [t5] dt + 162ʃ3 [t] dt – 42ʃ3 [t4] dt

Step 4: Now integrate the above expression with the help of power and trigonometry properties.

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 2 [t3 + 1 / 3 + 1]32 + 6 [t2 + 1 / 2 + 1]32 – 12 [t5 + 1 / 5 + 1]32 + 16 [t1 + 1 / 1 + 1]32 – 4 [t4 + 1 / 4 + 1]32

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 2 [t4 / 4]32 + 6 [t3 / 3]32 – 12 [t6 / 6]32 + 16 [t2 / 2]32 – 4 [t5 / 5]32

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 2/4 [t4]32 + 6/3 [t3]32 – 12/6 [t6]32 + 16/2 [t2]32 – 4/5 [t5]32

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 1/2 [t4]32 + 2 [t3]32 – 2 [t6]32 + 8 [t2]32 – 4/5 [t5]32

Step 5: Now use the fundamental theorem of calculus to apply the upper and lower limits of the integral.

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 1/2 [34 – 24] + 2 [33 – 23] – 2 [36 – 26] + 8 [32 – 22] – 4/5 [35 – 25]

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 1/2 [81 – 16] + 2 [27 – 8] – 2 [729 – 64] + 8 [9 – 4] – 4/5 [243 – 32]

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 1/2  + 2  – 2  + 8  – 4/5 

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 165/2 + 38 – 1330 + 40 – 844/5

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 82.5 + 38 – 1330 + 40 – 844/5

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 82.5 + 38 – 1330 + 40 – 168.8

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = 120.5 – 1330 + 40 – 168.8

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = -1209.5 + 40 – 168.8

2ʃ3 [2t3 + 6t2 – 12t5 + 16t – 4t4] dt = -1338.3

## Final Words

The integral is the fundamental branch of calculus that helps you to find the antiderivative, the area under the curve, and the volume. The types of the integral are used to find the integral with or without limits. The problems above will help to solve the integrals manually.