Application of Integrals

Integrals

In mathematics, Integration and Differentiation are the most significant ways to solve complex mathematical problems. Besides mathematics, Integration and Differentiation play key roles in Science, Engineering, and various other facets of our life. We have outlined what's, ifs, and how's related to integrals and their application. But how? Let us take an example.

To determine the area of a rectangle, the formulae 'length × breadth' is used. But what if you are given to calculate the area of the shaded portion, of a shaded rectangle with an unshaded circle within?

How is the Calculation Done? That is When Integrals Come Into the Play.

Integrals

In mathematics, the application of Integrals is applied to find the area under a curve, areas bounded by any curve, and so on.

Definition of Integrals

An integral is a function, of which a given function is a derivative. It is also known as the anti-derivative or reverse of a derivative.

Integrals are used to determine the area of 2D objects and the volume of 3D objects in real life.

Types of Integrals 

There are two types of Integrals. 

  • Definite Integrals

  • Indefinite Integrals

Definite Integrals:

An integral which has a start and an end value is known as a definite integral. In simple words, the function is restricted within an interval [a,b], where a and b are upper limit and lower limit, respectively. It is represented as \[y=\int_{b}^{a}f(x)dx\]

Example: y = ∫42 6x dx

Sol:  y= \[\int_{2}^{4}6x dx\]  here a = 4, b = 2

\[y = 6 [\frac{x^{2}}{2}]_{2}^{4} = 6[\frac{4^{2}}{2}] - [\frac{2^{2}}{2}] = 6[8-2]=36\]

Indefinite Integrals:

An integral which does not have an upper and lower limit is known as a definite integral. It is represented as 

\[y=\int_{b}^{a}f(x)dx\] = F(x)+C, where ‘C’ is a constant

Example: \[y=\int_{2}^{4}6x dx\]

Sol: \[y=\int 6xdx\]

\[y=6[\frac{x^{2}}{2}]\] = 3x + C, where C is a constant

Solved Examples

Question 1: Determine the Area Enclosed By a Circle  x2 + y2 =a2

Sol: It is observed that the area enclosed by the given circle is ‘4 x area of the region AOBA bounded by the curve, x-axis and the ordinates x=0 and x=a’.

[ as the circle is symmetrical about both x-axis and y-axis]

= \[4\int_{0}^{a}ydx\] (taking vertical strips)

=  \[4\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx\]

Since x2 + y2 =a2 gives \[y=\pm \sqrt{a^{2}-x^{2}}\]

The quarter AOBA lies in the first quadrant, hence 'y' is taken as positive. On integrating, we get the entire area enclosed by the given circle.

= \[4[[\frac{x}{2}]\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}]\] 

= \[4[([\frac{x}{2}]\times 0+\frac{a^{2}}{2}sin^{-1}1)-0]\]

= \[4(\frac{a^{2}}{2})(\frac{\pi}{2})\]

=\[\pi a^{2}\]


Question 2: Determine the Area of the Region Bounded b the Curve y = x2 and the line Y = 4

Sol: Since the given curve expressed by the equation y = x2is a parabola symmetric about y-axis only, therefore, the required area of the region AOBA is given bt 

= \[2\int_{0}^{4}x dy\]

= 2 area of the region BONB bounded by the curve, y-axis and the lines y=0 and y =4

= 2 x ( area of the region BONB bounced by the curve, y-axis and the lines y = 0 and y = 4 )

=\[2\int_{0}^{4}\sqrt{y}dy\]

= \[2\times\frac{2}{3}[y^{\frac{3}{2}}]_{0}^{4}\]

= \[\frac{4}{3}\times8\]

= \[\frac{32}{3}\]

Application of Integrals

Integrals have their application in both science and maths. In maths, the application of integral is made to determine the area under a curve, the area between two curves, the center of mass of a body, and so on. Whereas in science (Physics in particular), the application of integrals is made to calculate the Centre of Gravity, Mass, Momentum, Work done, Kinetic Energy, Velocity, Trajectory, and Thrust.

Application of Integrals in Engineering Fields:

There’s a vast application of integration in the fields of engineering.

In Architecture:

  • To determine the amount of material required in a curved surface. For instance, take the construction of a dome.

In Electrical Engineering:

  • Integrals are used in Electrical Engineering to calculate the length of a power cable required for transmission between two power stations.

Application of Integrals in Different Fields:

In Medical Science:

  • Integrals are used to determine the growth of bacterias in the laboratory by keeping variables such as a change in temperature and foodstuff.

In Medicine:

  • To study the rate of spread of infectious disease, the field of epidemiology uses medical seine to determine how fast a disease is spreading, its origin, and how to best treat it.

In Statistics:

  • To estimate survey data to help improve marketing plans for different companies because a survey requires many different questions with a range of possible answers.

FAQ (Frequently Asked Questions)

1. What is the Application of Integrals?

Integrals are broadly used to calculate areas under simple curves, the area between lines and arcs of circles, parabolas, and ellipses. It is seen how integration can be utilized to find an area between a curve and the x-axis. With very little modification, the area between a curve and the x-axis can be interpreted along with the area between the curve and a second curve. Additionally, Distance, Velocity, Acceleration, Volume, The average value of a function, Work, Center of Mass, Kinetic energy, improper integrals, Probability, Arc Length, Surface Area can be defined with equations; the idea is quite easy to understand.

With this lesson, we hope the concept of integrals is clear. Likewise, we can apply integrals to find the area of an enclosed figure, the area of an enclosed area bounded by the curve, or by the x-axis and y-axis.

2. What is the Relationship of Integration and Differentiation?

Integration and differentiation are related to each other just like square and square root functions. Say for instance we take a number x and square it we get X2 . Now if we operate the square root function on X2 we get the original  x back. Similarly when we operate integration on a continuous function F we get a new integral. If we were to operate differentiation on this new integral we will get the original function F at the end.