Integral Calculus: Definition, Types, Properties, and Calculations

The term integral is a well-known branch of calculus in which the study of integration and its properties is under consideration. There are two main branches of calculus such as integral calculus and differential calculus.

Differential calculus is the rate of change of any algebraic or trigonometric functions with respect to the independent variable. While the integral is the reverse or anti-process of the derivative. In this post, we are going to discuss integral calculus along with solved examples.

What Is Integral Calculus?

The term integral is the building block of calculus along with differentiation. The integral calculus is used to determine the area under the curve. This area could be a numerical value or an expression. 

The integral calculus can be evaluated easily if the differential f’(y) of the function is given. In differential calculus, we take the function f(y) and determine the f’(y) of the function. While in integral calculus we have to take the f’(y) as a function to determine f(y) (original function).

The integral calculus is usually the inverse of determining differential. This process of finding the original function from f’(y) is said to be the anti-derivative, anti-differentiation, or integration. 

Types of Integration

The integral or anti-derivative is further divided into two types such as definite integral and indefinite integral. 

Definite integral	Indefinite integral
The definite integral is a type of integration that deals with the numerical value of the function.	The indefinite integral is the other type of integration in which the new function has to be determined whose original function is a differential f’(y).
The upper and lower limit values are used to evaluate the numerical value of the function. The limit values will lie on the real line. And the values of the start and end point of the function will be applied with the help of the fundamental theorem of calculus	There is no boundary or limit value used in this type of integral as it is the reverse process of differentiation to evaluate the new function.
The constant of integration is not applied in this type of integral.	The constant of integration is applied in this type of integral.
It is also known as Riemann integral.	It is also known as anti-derivative.

Formulas of Integral Calculus

Below are the basics integral formulas for definite and indefinite integrals.

Definite integral formula	Indefinite integral formula
The general formula for finding the definite integral is:∫ba f(y) dy = [F(y)]ba = F(b) – F(a)	The general formula for finding the indefinite integral is:∫ f(y) dy = F(y) + C
f(y) is the differential function.dy is the integrating variable of the functiona & b are the limit values of the function.F(y) is the integral of the functionF(b) – F(a) is the fundamental theorem of calculus for applying limits.	f(y) is the differential function.dy is the integrating variable of the functionF(y) is the integral of the functionC is the constant of integration

Properties of Integral Calculus

Here are a few properties of integral calculus.

• The integrand of the integral is the derivative f’(y).
• The integral of two summed functions is equal to the sum of the integral function separately.

∫ [ f(y) dy + g(y) dy] = ∫ f(y) dy + ∫ g(y) dy

• Similarly, the integral of two difference functions is equal to the difference of integral functions separately.

∫ [ f(y) dy – g(y) dy] = ∫ f(y) dy – ∫ g(y) dy

• The constant coefficient along with the function should be taken outside the integral notation. 

∫ k f(y) dy = k ∫ f(y) dy

Calculations of Integral Calculus

Here we’ll discuss how to perform the calculations of complex integral functions. The formulas and rules of integral calculus will be helpful for calculating integral calculus. Below are a few examples of integral calculus.

Example 1: For indefinite integral

Find the antiderivative of the given differential function. 

f(y) = 4y + 3y³ – 4y⁵ + 2y⁵ + 20

Solution 

Step 1: First of all, take the given differential function and write it as an integrand of integral calculus. 

f(y) = 4y + 3y³ – 4y⁵ + 2y⁵ + 20

ʃ f(y) dy = ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy

Step 2: Now use the properties of the integral calculus to the above expression to write the notation of integral with each function separately.

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = ʃ [4y] dy + ʃ [3y³] dy – ʃ [4y⁵] dy + ʃ [2y⁵] dy + ʃ [20] dy

Step 3: Now use the last property of taking constant outside the integral notation to the above expression.   

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 4ʃ [y] dy + 3ʃ [y³] dy – 4ʃ [y⁵] dy + 2ʃ [y⁵] dy + ʃ [20] dy

Step 4: Now find the integral of the above expression.

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 4 [y¹⁺¹ / 1 + 1] + 3 [y³⁺¹ / 3 + 1] – 4 [y⁵⁺¹ / 5 + 1] + 2 [y⁵⁺¹ / 5 + 1] + [20y] + C

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 4 [y² / 2] + 3 [y⁴ / 4] – 4 [y⁶ / 6] + 2 [y⁶ / 6] + [20y] + C

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 4/2 [y²] + 3/4 [y⁴] – 4/6 [y⁶] + 2/6 [y⁶] + [20y] + C

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 2 [y²] + 3/4 [y⁴] – 2/3 [y⁶] + 1/3 [y⁶] + [20y] + C

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 2y² + 3y⁴/4 + [-2/3 + 1/3] y⁶ + 20y + C

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 2y² + 3y⁴/4 + [-1/3] y⁶ + 20y + C

ʃ [4y + 3y³ – 4y⁵ + 2y⁵ + 20] dy = 2y² + 3y⁴/4 – y⁶/3 + 20y + C

Try any integral calculator to check the results of the calculated problems. 

The above problem is solved through integral calculator by Meracalculator.

Example 2: For definite integral 

Find the integral of the given differential function by using upper and lower limits such as [1, 2]. 

f(x) = 3x – 2x² + 6x² – 5x³ + 2x

Solution 

Step 1: First of all, take the given differential function and write it as an integrand of integral calculus. 

f(x) = 3x – 2x² + 6x² – 5x³ + 2x

ʃ^b_a f(x) dx = ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx

Step 2: Now use the properties of the integral calculus to the above expression to write the notation of integral with each function separately.

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = ʃ²₁ [3x] dx – ʃ²₁ [2x²] dx + ʃ²₁ [6x²] dx – ʃ²₁ [5x³] dx + ʃ²₁ [2x] dx 

Step 3: Now use the last property of taking constant outside the integral notation to the above expression.   

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3ʃ²₁ [x] dx – 2ʃ²₁ [x²] dx + 6ʃ²₁ [x²] dx – 5ʃ²₁ [x³] dx + 2ʃ²₁ [x] dx 

Step 4: Now find the integral of the above expression.

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3 [x¹⁺¹ / 1 + 1]²₁ – 2 [x²⁺¹ / 2 + 1]²₁ + 6 [x²⁺¹ / 2 + 1]²₁ – 5 [x³⁺¹ / 3 + 1]²₁ + 2 [x¹⁺¹ / 1 + 1]²₁

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3 [x² / 2]²₁ – 2 [x³ / 3]²₁ + 6 [x³ / 3]²₁ – 5 [x⁴ / 4]²₁ + 2 [x² / 2]²₁

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3/2 [x²]²₁ – 2/3 [x³]²₁ + 6/3 [x³]²₁ – 5/4 [x⁴]²₁ + 2/2 [x²]²₁

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3/2 [x²]²₁ – 2/3 [x³]²₁ + 2 [x³]²₁ – 5/4 [x⁴]²₁ + [x²]²₁

Step 4: Now with the help of the fundamental theorem of calculus apply the limits values of the function.

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3/2 [2² – 1²] – 2/3 [2³ – 1³] + 2 [2³ – 1³] – 5/4 [2⁴ – 1⁴] + [2² – 1²]

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3/2 [4 – 1] – 2/3 [8 – 1] + 2 [8 – 1] – 5/4 [16 – 1] + [4 – 1]

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 3/2 [3] – 2/3 [7] + 2 [7] – 5/4 [15] + [3]

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 9/2 – 14/3 + 14 – 75/4 + 3

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = 4.5 – 4.67 + 14 – 18.75 + 3

ʃ²₁ [3x – 2x² + 6x² – 5x³ + 2x] dx = -1.92

Final Words

Now you are able to take any basic concept from this post. We have discussed Integral Calculus, its types, formulas, properties, and calculations. Going through this post and practicing them by hand will make you master in this topic.