In mathematics, the integral is a branch of calculus used to find the integrals of the functions. The area of the region can be calculated by using this type of calculus along with the graph of the function. The term differentiation is related to integration.
The reverse process of the differential is known as the integral. These branches of mathematics are very essential to evaluate the complex problems of mathematics. In this post, we will learn the definition, types, and calculations of integration.
What is the integral?
Integration is a branch of mathematics used to explain the applications and theory for the solution of the differential equations to determine the areas, volumes, and lengths. The process of finding the integral of the function is known as integration.
The antiderivative is also used in place of integral. In simple words, the other name of integral is antiderivative. The integral is used to integrate the function with or without applying the limits.
Types of integral
There are two main types of integrals used to evaluate the problems of calculus with or without applying the limits.
- Definite integral
- Indefinite integral
Let us describe the types of integration briefly.
Definite integral
The definite integral is that type of integration in which the upper and lower limits exist. The start and endpoint of the function by applying the integral notation is known as the definite integral. When the values of the integral are lies on the real line then the definite integral can also be named as a Riemann integral.
The notation of the definite integral is:
f(x) dx = F(x)
- f(x) is the function.
- dx is the respective variable.
- a and b are the lower and upper limits respectively.
- ʃ is the integral notation.
- F(x) is the integrated result of the function.
After integrating the function, the limits can be substituted in a function by applying the fundamental theorem of calculus.
f(x) = F(b) – F(a)
Indefinite integral
The indefinite integral is the other type of integral in which limits are not applied. In simple words, the upper and lower limits are not used in this type of integral. The indefinite integral reverses the process of the differential function to returns a function of the independent variable.
The equation of the indefinite integral is:
ʃ f(x) dx = F(x) + C
- f(x) is a function to be integrated known as the integrand.
- dx is the integrating variable.
- C is the constant of integration.
- x is the integration variable.
- F(x) is the integral or antiderivative of the function.
According to the above equation, the antiderivative of f(x) is F(x) with respect to x.
You can use an integral calculator to get the result according to the types of integral.
How to calculate the problems of integral?
By using the definite and indefinite integral, you can easily solve the problems of integration. Let us take some examples
Example 1: For indefinite integral
Calculate 4x2 + 5sin(x) – 2x4y + x – tan(x) using indefinite integral, with respect to x.
Solution
Step 1: Write the integral notation with the given function.
ʃ (4x2 + 5sin(x) – 2x4y + x – tan(x)) dx
Step 2: Apply the sum and difference rules in the above integral function.
ʃ (4x2 + 5sin(x) – 2x4y + x – tan(x)) dx = ʃ (4x2) dx + ʃ 5sin(x) dx – ʃ 2x4y dx + ʃ x dx – ʃ tan(x) dx
Step 3: Write the constants outside the integral notation.
ʃ (4x2 + 5sin(x) – 2x4y + x – tan(x)) dx = 4ʃ (x2) dx + 5ʃ sin(x) dx – 2 y ʃ x4 dx + ʃ x dx – ʃ tan(x) dx
Step 4: Now integrate the above equation by apply the power and trigonometric rules.
ʃ (4x2 + 5sin(x) – 2x4y + x – tan(x)) dx = 4 (x2+1 / 2 + 1) + 5 (-cos(x)) – 2 y (x4+1 / 4 + 1) + (x1+1 / 1 + 1) – (-ln(cos(x)) + C
ʃ (4x2 + 5sin(x) – 2x4y + x – tan(x)) dx = 4 (x3 / 3) + 5 (-cos(x)) – 2 y (x5 / 5) + (x2 / 2) – (-ln(cos(x)) + C
ʃ (4x2 + 5sin(x) – 2x4y + x – tan(x)) dx = 4x3 / 3 – 5cos(x) – 2yx5 / 5 + x2 / 2 + ln(cos(x) + C
ʃ (4x2 + 5sin(x) – 2x4y + x – tan(x)) dx = 4/3 (x3) – 5cos(x) – 2/5 (x5y) + 1/2 (x2) + ln(cos(x) + C
Example 2: For definite integral
Evaluate the given function by using definite integral having upper and lower limits 4 and 3, 18x3 + 23sin(x) – 25x3y2 + 5x
Solution
Step 1: Write the integral notation with the given function along with the upper and lower limit.
(18x3 + 23sin(x) – 25x3y2 + 5x) dx
Step 2: Apply the sum and difference rules on the above integral function.
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = (18x3) dx + (23sin(x)) dx – (25x3y2) dx + (5x) dx
Step 3: Write the constants outside the integral notation.
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 18 (x3) dx + 23 (sin(x)) dx – 25y2 (x3) dx + 5 (x) dx
Step 4: Now integrate the above equation by apply the power and trigonometric rules.
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 18(x3+1/3 + 1)43 + 23(-cos(x))43 – 25y (x3+1/3 + 1)43 + 5(x1+1/1 + 1)43
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 18(x4/4)43 + 23(-cos(x))43 – 25y (x4/4)43 + 5(x2/2)43
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 18/4 (x4)43 – 23(cos(x))43 – 25y/4 (x4)43 + 5/2 (x2)43
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 9/2 (x4)43 – 23(cos(x))43 – 25y/4 (x4)43 + 5/2 (x2)43
Step 5: Now apply the fundamental theorem of calculus f(x) = F(b) – F(a) and put the limits in the above equation.
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 9/2 (44 – 34) – 23(cos (4) – cos (3)) – 25y/4 (44 – 34) + 5/2 (42 – 32)
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 9/2 (256 – 81) – 23(cos (4) – cos (3)) – 25y/4 (256 – 81) + 5/2 (16 – 9)
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 9/2 (175) – 23(cos (4) – cos (3)) – 25y/4 (175) + 5/2 (7)
(18x3 + 23sin(x) – 25x3y2 + 5x) dx = 1575/2 – 23cos (4) + 23cos (3) – 4375y/4 + 35/2
Summary
Now you are witnessed that the integral calculus topic is not much difficult. Once you grab the basics of this post, you can easily solve the problems of integration without any difficulty. You can also use an integral solver to verify your results.
