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
∫ab g(y) dy = |G(y)|ab = 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.
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
Example 2: For indefinite
Find the Indefinite Integrate of the function a * (x2 + b) + x5 – y where the variable of integration is x.
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
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.
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)
Graph: The area under the given curve shown below
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.