What is Integral? Explained with its types, applications, and graphs
Integrals are used in many domains,
including mathematics, science, and engineering. We mostly utilize integral
formulae to calculate areas.
So, let us offer a quick
introduction to integrals based on the Mathematics topic in order to determine
areas under simple curves, areas limited by a curve and a line, and areas
between two curves, as well as the application of integrals in the mathematical
disciplines and the solved problem.
What
is an Integral?
Integral is the reverse of derivative. The answer
of the integral is the function whose derivative is given. The process of
finding the integral is known as integration. An integral is frequently used to evaluate the area under the curve.
Integration is also known as the anti-derivative
of the function.
The
integral formulae
The formulae of the integral are used to
integrate the given function with or without limits.
The notation of integral is
Êƒ
f(x) dx
•
f(x) is the
function.
•
dx is the
integral variable of integration with respect we take the integral.
•
Êƒ the sign of integral.
Types
of Integral
The integral (anti-derivative) has two types.
•
Indefinite
integral
•
definite
integral
We discuss
this one by one in detail with examples.
•
Indefinite
integral
The indefinite integral is an integral of a
function without using the upper and lower limits. it is represented by
Êƒ f(x) dx =
F(x) +C
•
F(x) is the
function after taking integral
•
C is called
the constant of integration
•
f(x) is the
function whose integral we have to find
•
dx is
called the integral variable
•
Definite
integral
The definite integral is another type of
integral. In this integral lower limit and upper limit involves the integral
sign.
Êƒ^{b}_{a}
f(x) dx = F(b) – F(a)
•
f(x) is
a function.
•
a and b is called the lower limit and upper
limit of the function respectively.
•
dx is known
as the integral variable with respect to the procedure of integration carry.
•
No constant
involves
•
We put the
upper limit first and then minus the lower limit.
Examples
of integral calculus
We discuss the each type with the help of
examples and detailed solutions with step-by-step elaboration.
Example 1:
Evaluate the definite integral of f(x) = sin(5x)
+ x^{2 }– 5x where x is an integrating variable and the variation of x
is from 2 to 3.
Solution:
Step
1: Change into the
integral notation.
Êƒ^{3}_{2} sin (5x) + x^{2 }–
5x) dx
Step
2: Apply the integral
separately on each term.
= Êƒ^{3}_{2} [sin (5x)] dx + Êƒ^{3}_{2}
[x^{2}] dx^{ }– Êƒ^{3}_{2} [5x] dx
Step
3: Taking the
coefficient outside of the integral sign.
= Êƒ^{3}_{2} [sin (5x)] dx + Êƒ^{3}_{2}
[x^{2}] dx^{ }– 5Êƒ^{3}_{2} [5x] dx
Step
4: Now apply the integral
= [-cos (5x)/5]^{3}_{2} + [x^{2+1}
/ 2 + 1]^{3}_{2}^{ }– 5 [x^{1+1} /1 + 1]^{3}_{2}
= [-cos (5x)/5]^{3}_{2} + [x^{3}
/ 3]^{3}_{2}^{ }– 5 [x^{2} /2]^{3}_{2}
= -1/5[cos (5x)]^{3}_{2} + 1/3 [x^{3}]^{3}_{2}^{
}– 5/2 [x^{2}]^{3}_{2}
Step
6: Put the limits
according to the fundamental theorem of calculus.
= -1/5[cos (5(3)) – cos (5(2))] + 1/3 [3^{3}
– 2^{3}]^{ }– 5/2 [3^{2} – 2^{2}]
= -1/5[cos (15) – cos (10)] + 1/3 [27 – 8]^{ }–
5/2 [9 – 4]
= -1/5[cos (15) – cos (10)] + 1/3 [19]^{ }–
5/2 [5]
= -1/5[cos (15) – cos (10)] + 19/3^{ }–
25/2
= -1/5[cos (15) – cos (10)] + 37/6
= -6.1825
The calculation of the integral is sometimes a
very difficult and time taking procedure. To avoid these difficulties try
online integral calculators.
Solved through integral calculator by Meracalculator
Example 2:
Find the
area under the given function √(16 – x^{2}).
Solution:
Step
1: Find the limits
√(16 – x^{2}) = 0
Taking the square on both sides
16 – x^{2}
= 0
4^{2} – x^{2} = 0
(4-x) (4 + x) = 0
4
- x = 0 |
4+x
= 0 |
x = 4 |
x = -4 |
So the limits of integration are 4, -4
Step
2: Now write in the
form of the definite integral.
Êƒ^{4}_{-4}√(16 – x^{2})dx
Step
3: Draw the graph of
the given function.
Step
4: The graph of the
integrated function with limits -4 to 4 as the semi-circle with a radius of 4.
Use the formula of the area of the semi-circle
Area of semi-circle = (pi*r^2)/2
Step
5: Put the value of
the radius.
Area of the semi-circle = (3.14*4^{2})/2
Area of the semi-circle = 25.12
Example 3:
Evaluate the indefinite integral of the function
2x^{3 }– 9x^{2 }+ 5cos(10x) with integrate variable x.
Solution:
Step
1: Change into the
integral notation.
Êƒ [2x^{3}- 9x^{2}+5cos (10x)] dx
Step
2: Apply the integral
according to the rule that it separates on each term
= Êƒ (2x^{3}) dx – Êƒ (9x^{2}) dx +
Êƒ 5cos (10x)) dx
Step
3: Take the
coefficient outside the integral sign.
= 2Êƒ (x^{3}) dx – 9Êƒ (x^{2}) dx +
5Êƒ cos (10x)) dx
Step
4: Apply integral now
= 2 (x^{3+1}/ 3+1) – 9 (x^{2+1}/
2+1) + 5[sin (10x)/10] + C
= 2 (x^{4}/4) – 9 (x^{3}/ 3) +
5[sin (10x)/10] + C
= 2/4 (x^{4}) – 9/3 (x^{3}) +
5/10[sin (10x)] + C
= 1/2 (x^{4}) – 3 (x^{3}) +
1/2[sin (10x)] + C
= x^{4}/2 – 3x^{3} + sin(10x)/2 +
C
Application
of integral in real life
The application of integrations in real life is
determined by the industries in which this calculus is employed. We discuss
some real-life applications the integral is following
•
Engineers,
for example, utilize integrals to determine the geometry of building projects
or the length of electrical wire necessary to connect two substations.
•
It is used
in science to address questions about the center of gravity and other physics
concerns.
•
In the field of graphical representation,
three-dimensional models are displayed.
•
Kinetic
energy and improper integrals
•
Probability
theory
•
The area
between the curve
Summary
In this post, we have learned about integration,
its types, examples, and real-life applications.
You can now able to solve both types of integral
questions easily and with better accuracy. The integral is one of the main branches
of calculus.