Types of Integral: Defined and Explained with Calculations

Numerous fields, such as mathematics, physics, and engineering, use integrals. For the most part, when calculating areas, we use integral formulas.

In order to determine the areas under simple curves, the areas bounded by a curve and a line, and the areas between two curves, as well as the application of integrals in the mathematical disciplines and the solved problem, let us first provide a brief introduction to integrals based on the topic of mathematics.

What is an Integral?

Derivative is the opposite of integral. The function whose derivative is given is the integral’s solution. Integration is the practice of determining the integral. The anti-derivative of the function is another name for integration.

The integral formulae

The formulae of the integral are used to integrate the given function with or without limits.

The notation of integral is

ʃ g(y) dy

g(y) is the function.
dy is the integral variable of integration with respect we take the integral.
ʃ the sign of integral.

Types of integral:

There are two types of integral we use in integral calculus.

Definite integral
Indefinite integral

Let us briefly describe the types of integral.

1. Indefinite integral:

The method that is used to find a new function whose original function is differential is said to be the indefinite integral. The indefinite integral of a function g(y) is defining as

∫ g(y) dy = G(y) + C

G(x) the function after the integral we obtain
g(x) the function whose integral has been taken
C is the constant of integral
∫ in the integral sign
dy is the differential of the variable y.

2. Definite integral:

The method used to calculate the numerical values of the function by taking the boundary points is said to be the definite integral. The boundary points can be substituted with the help of the fundamental theorem of calculus.

The definite integral of a function g(y) is defining as

∫^a_b g(y) dy = |G(y)|^a_b = G(a) – G(b)

a is known as the upper limit
b is called lower limit
g(y) is called the integrand.
G(y) is the function after integration

An integral calculator is a helpful tool to calculate the definite and indefinite integral with steps.

Examples of integral

We will discuss some examples to understand the both definite and indefinite integral with step-by-step solution.

Example 1: For definite

Evaluate the definite integralcos (5y) +y³-6y) where y is an integrating variable and the variation of y is from 2 to 3.

Solution:

Step 1: First we write into the integral notation.

= cos (5y) +y³-6y) dy

Step 2: Apply the integral on each term one by one

= cos (5y) dy +y³ dy-6y dy

Step 3: Taking the coefficients of the variables outside the integral sign.

= cos (5y) dy + y³ dy- 6 y dy

Step 4: Now apply the integral

= (sin(5y) / y)³₂ + (y⁴/4)³₂ -5 (y² /2)³₂

Step 5: Take the constant term outside the limit’s notation

= 1/5 (sin(5y))³₂ + 1/4 (y⁴)³₂ – 5/2 (y²)³₂

Step 6: Put the limits according to the rule

= 1/5 (sin(5*3) – sin( 5 *2)) + 1/4 (3⁴ – 2⁴) – 5/2( 3² – 2²)

= 1/5 (sin(15) – sin (10) + 1/4 (81 – 16) – 5/2(9 – 4)

= 1/5 (sin(15) –sin( 10)) + ¼ (65) – 5/2*(5)

= 1/5 (-cos 15 + cos 10) + 65/4 – 25/2

= 1/5 (cos 10 – cos 15) + (65 – 75)/6

= 1/5 (cos 10 – cos 15) – 10/6

= 0.4889

Example 2: For indefinite

Find the Indefinite Integrate of the function a * (x² + b) + x⁵– y where the variable of integration is x.

Solution:

Step 1: Write the function in the form of integral notation:

= ∫ [a * (x² + b) + x⁵– y] dx

Step 2: Apply the integrate separately one by one on every term.

= ∫a * (x² + b) dx + ∫ x⁵ dx – ∫ y dx

Step 3: Now factors out the constant’s terms.

= a ∫ x² dx + ∫ x⁵ dx – (ab – y) ∫ 1 dx

Step 4: Now integrate the terms

= a x³/3 + x⁶/6 – (ab-y) x + C

= abx + ax³/3 + x⁶/6 -xy + C

Example 3:

Find the definite integral x³/ (x²+ 1) where the upper limit is 3 and lower limit is 2 where the variable of integral is x.

Solution:

Step 1: Write the expression in the definite integral notation

= ∫³₂(x³/ (x² + 1)) dx

Step 2: We use substitution method for integral

Substitute u = x² and du = 2x dx

For lower limit u = 2² = 4 and upper limit is u = 3² = 9:

= ∫⁹₄(u/ (u + 1)) du

Step 3: For the integrand u/(u+1), do long division:

= 1/2 ∫⁹₄(1 – 1/ (u + 1)) du

Step 4: Integrate the sum term by term and factor out constants:

= 1/2 ∫⁹₄1 du – 1/2 ∫⁹₄(1/ (u + 1)) du

Step 5: For the integrand 1/1+u, substitute s = u+1 and ds = du.

This gives a new lower limit s= 1+4 =5 and upper limit s = 1+9 = 10

= 1/2 ∫⁹₄1 du – 1/2 ∫¹⁰₅(1/ (s) ds

Step 6: Apply the fundamental theorem of calculus:

The anti-derivative of 1/s is log(s)

= |(-log(s)/2|¹⁰₅ + 1/2|u|⁹₄

Step 7: Evaluate the antiderivative at the limits and subtract:

= (-log (10)/2) -( (-log (5)/2) + 1/2(9-4)

= -log (2)/2 + 5/2

= 1/2 (5 – log (2))

= 1/2 (5 – 0.3010)

=1/2 (4.69897)

= 2.3494

Graph: The area under the given curve shown below

Summary:

In this post, we encounter with the integral, its types, methods and examples. You can now find answers of both kinds of integral questions quickly and accurately.

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