History of the Integral from the 17th Century & all things about integral from daily life application to its formula

1.1 Introduction

The path to the development of the integral is a branching one, where similar discoveries were made simultaneously by different people. The history of the technique that is currently known as integration began with attempts to find the area underneath curves. The foundations for the discovery of the integral were first laid by Cavalieri, an Italian Mathematician, in around 1635. Cavalieri’s work centered around the observation that a curve can be considered to be sketched by a moving point and an area to be sketched by a moving line.

1.2 Cavalieri's Method of Indivisbles
In order to deal with the geometrical notion of a moving point, Cavalieri worked with what he called "indivisibles". That is, if a moving point can be considered to sketch a curve, then Cavalieri viewed the curve as the sum of its points. By this notion, each curve is made up of an infinite number of points, or "indivisibles". Likewise, the "indivisibles" that composed an area were an infinite number of lines. Though Cavalieri was not the first person to consider geometric figures in terms of the infinitesimal (Kepler had done so before him), he was the first to use such a notion in the computation of areas (Hooper 248-250).
In order to introduce Cavalieri’s method, consider finding the area of a triangle. For many years, it had been known that the area of a triangle was ½ the area of a rectangle which has the same base and height.

Cavalieri now took a step of great importance to the formation of the integral calculus. He utilized his notion of "indivisibles" to imagine that there were an infinite number of shaded regions. He saw that as the individual shaded regions became small enough to simply be lines, the jagged steps would gradually define a line. As the jagged steps became a line, the shaded region would form a triangle. As the number of shaded regions increases, the ratio remains simply one-half.
Cavalieri’s methodology agreed with the long-held result that the area of a triangle was one-half the product of the base and height. He had also shown that his notion of "indivisibles" can be used to successfully describe the area underneath the curve. That is, as the areas of the rectangles turn into lines, their sum does indeed produce the area underneath the curve (in this case, a line). Cavalieri went on to use his method of "indivisibles" to find the area underneath many different curves. However, he was never able to formulate his techniques into a logically consistent foundation that others accepted. Though Cavalieri’s techniques clearly worked, it was not until Sir John Wallis of England that the limit was formally introduced in 1656 and the foundation for the integral calculus was solidified (Hooper 249-253).
In order to fully understand Wallis’ contributions to the integral calculus, it is first necessary to see how Cavalieri’s theoretical techniques can be applied to find the area underneath a curve more complicated than a line. In order to do so, this technique will be applied to find the area underneath the parabola.
Each rectangular region has a base of 1 unit along the x-axis and height of x²(obtained from the definition of the parabola). The number of rectangular regions will be defined to be m. Cavalieriagain attempted to express the area underneath the curve as the ratio of an area that was already known. He considered the area enclosing all of the mrectangles. It can easily be seen from the diagram that the base of this rectangle will be m+1 (there are m rectangles, the first starting at ½ and the last one ending at m + ½). The height of the enclosing rectangle will be m², from the definition of the parabola. The ratio can now be expressed with the following equation:

Recall that the area of a rectangle is defined by the product of its base and height. It was stated that the bounding rectangle had a base of m+1 and a height of m², which accounts for the denominator. The numerator is easily explained as well: each of the m rectangles has a base of 1 and a height of its x value squared. Cavalieri now proceeded to calculate the ratio for different values of m. In doing so, he noticed a pattern and was able to establish a closed form for the ratio of the areas:

Cavalieri then utilized his important principle of "indivisibles" to make another important leap in the development of the calculus. He noticed that as he let m grow larger, the term had less influence on the outcome of the result. In modern terms, he
noticed that

That is, as he lets the number of rectangles grow to infinity, the ratio of the areas will become closer to

. Though Cavalieri did not formally introduce the notation for limits, he did utilize the idea in the computation of areas. After using the concept of infinity to describe the ratios of the area, he was able to derive an algebraic expression for the area underneath the parabola. For at any distance x along the x-axis, the height of the parabola would be x². Therefore, the area of the rectangle enclosing the rectangular subdivisions at a point x was equal to or x³. From his earlier result, the area underneath the parabola is equal to 1/3 the area of the bounding rectangle. In other words:

With this technique, Cavalieri had laid the fundamental building block for integration.

1.3 Wallis' Law for Integration of Polynomials
John Wallis’ contribution to the integral calculus was to derive an algebraic law for integration that alleviated the necessity of going through such analysis for each curve. Through examining the relationship between a function and the function that describes its area (henceforth referred to as the area-function), he was able to derive an algebraic law for determining area-functions. Rather than simply present the algebraic relationship (which the reader is doubtless familiar with if (s)he has studied a minimal amount of calculus), we will perform a similar analysis as to what led Wallis to derive his law.

y = k

Clearly, it can be seen from the diagram, that the area underneath the line at any point along the x-axis will be kx or .
Next, consider the graph of the function y = kx :

At any point x along the x-axis, the height will be equal to kx. Since the area forms a triangle, the area underneath the curve can be expressed as ½ the base times the height or . As was already shown above, the area underneath a parabola
y = kx², can be expressed as . Wallis noticed an algebraic relationship between a function and its associated area-function. That is, the area-function of is . Wallis went on to show that not only does this hold true where n is a natural number (which had been the extent of Cavalieri’s work), but that it also worked for negative and fractional exponents. Wallis also showed that the area underneath a polynomial composed of terms with different exponents (e.g. ) can be computed by using his law on each of the terms independently (Hooper 255 - 260).

1.4 Fermat's Approach to Integration
One of the first major uses of infinite series in the development of calculus came from Pierre De Fermat’s method of integration. Though previous methods of integration had used the notion of infinite lines describing an area, Fermat was the first to use infinite series in his methodology. The first step in his method involved a unique way of describing the infinite rectangles making up the area under a curve.

Figure 1.5 Fermat noticed that by dividing the area underneath a curve into successively smaller rectangles as x became closer to zero, an infinite number of such rectangles would describe the area precisely. His methodology was to choose a value 0 < e < 1, such that a rectangle was formed underneath the curve at each power of e times x (see Figure 1.5, NOTE: e was simply Fermat’s choice of variable names, not e = 2.71828…). Fermat then computed each area individually:

The first equation represents the area of the largest rectangle, the second equation the next rectangle to the left, and so on. The areas are simply found by multiplying the base times the height. The base is known by the power of e, and the height by evaluating at the given x value. The simplifications of each area expression are given in a form that will be useful when attempting to find the infinite sum.Fermat’s next step was to compute the infinite sum of these rectangles as the power of e approached infinity.

By determining the sum of each increasing finite series, he was able to develop an expression for the infinite sum.
In order to find a closed form for the expression

...note that the sum is a geometric series of the form:

If 0 < x < 1, the sum is (this can be shown to be true by long dividing (1-x) into 1). Therefore, by substituting back in for x and inserting into the overall equation, the area can be expressed as:

Fermat now wished to express the area entirely in terms of x, and in order to do so substituted , which by simplification and factoring out (1-E):

Fermatll now made a step that with the benefit of current knowledge is explainable, but at that time was not properly justified. That is, Fermat said let E = 1 and since and because then e must also equal 1. By substituting 1 for E in the area expression above:

Although this methodology yielded the appropriate result for the area underneath the curve, Fermat’s justification of letting E = 1 was not properly formulated. What he actually was doing was taking the limit as E approaches 1 and as E approaches 1 so too will e. As e approaches 1, then e raised to any power will also approach 1, and the infinite sum of the areas underneath the curve has been determined. The notion of a limit was hinted at in Fermat’s work, but it was not formally defined until later (Boyer 162 - 169).
Wallis and Fermat's work had laid the groundwork for the modern concept of the integral. However, what Fermat and Wallis had failed to recognize was the relationship between the differential and the integral. That idea would be developed simultaneously by two men: Newton and Leibniz. This would later be known as the Fundamental Theorem of Calculus and, as the National implies, it is a landmark discovery in the history of the Calculus. However, before proceeding on to describe this important theorem, it is first necessary to examine the development of the differential.

Application of calculus.

It might seem like an age-old question, but have you ever pondered over “When am I ever going to use this in real life” in your math class? To stumble on the applications of basic arithmetic or finances are obvious, but when it comes to calculus, we often find ourselves scratching our heads with confusion.

There’s one thing I need to tell you: Of course, you may not sit down and solve a tricky differential equation on a daily basis, but calculus is still all around you. Did you know that because of the ability to model and control systems, it gives us extraordinary power over the material world?

Calculus is the language of engineers, scientists, and economists. The work of these professionals has a huge impact on our daily life – from your microwaves, cell phones, TV, and car to medicine, economy, and national defense.

What is Calculus?

British Scientist Sir Isaac Newton (1642-1727) invented this new field of mathematics. ‘Calculus’ is a Latin word, which means ‘stone.’ Romans used stones for counting. For the counting of infinitely smaller numbers, Mathematicians began using the same term, and the name stuck.

Though it was proved that some basic ideas of Calculus were known to our Indian Mathematicians, Newton & Leibnitz initiated a new era of mathematics. There are a large number of applications of calculus in our daily life. It can’t be possibly illustrated in a single view. I’m trying to describe them shortly here.

Application in Engineering

An Architect Engineer uses integration in determining the amount of the necessary materials to construct curved shape constructions (e.g. dome over a sports arena) and also to measure the weight of that structure. Calculus is used to improve the architecture not only of buildings but also of important infrastructures such as bridges.
In Electrical Engineering, Calculus (Integration) is used to determine the exact length of power cable needed to connect two substations, which are miles away from each other.
Space flight engineers frequently use calculus when planning for long missions. To launch an exploratory probe, they must consider the different orbiting velocities of the Earth and the planet the probe is targeted for, as well as other gravitational influences like the sun and the moon.

Application in Medical Science

Biologists use differential calculus to determine the exact rate of growth in a bacterial culture when different variables such as temperature and food source are changed.

Application in Physics

In Physics, Integration is very much needed. For example, to calculate the Centre of Mass, Centre of Gravity and Mass Moment of Inertia of a sports utility vehicle.
To calculate the velocity and trajectory of an object, predict the position of planets, and understand electromagnetism.

Application in Statistics

Statisticians use calculus to evaluate survey data to help develop business plans for different companies. Because a survey involves many different questions with a range of possible answers, calculus allows a more accurate prediction for the appropriate action.

Application in Research Analysis

An operations research analyst will use calculus when observing different processes at a manufacturing corporation. By considering the value of different variables, they can help a company improve operating efficiency, increase production, and raise profits.

Application in Graphics

A graphics artist uses calculus to determine how different three-dimensional models will behave when subjected to rapidly changing conditions. It can create a realistic environment for movies or video games.

Application in Chemistry

It is used to determine the rate of a chemical reaction and to determine some necessary information of Radioactive decay reaction.

Let’s take a look at some other examples.

Credit card companies use calculus to set the minimum payments due on credit card statements at the exact time the statement is processed by considering multiple variables such as changing interest rates and a fluctuating available balance.
Doctors and lawyers use calculus to help build the discipline necessary for solving complex problems, such as diagnosing patients or planning a prosecution case.
The field of epidemiology — the study of the spread of infectious disease — relies heavily on calculus. It can be used to determine how far and fast a disease is spreading, where it may have originated from and how to best treat it.

For most professions, this topic is very important. So, this is why you can’t dismiss it as just another subject. Despite its mystique as a more complex branch of mathematics, it touches our lives each day, in ways too numerous to calculate.
You can find tips from our experts here to become super at calculus.

Integration: Formulas, Definite & Indefinite Integral

Integration or Integral Calculus is often lightly passed upon the students as the reverse operation of Differential Calculus, which is true but doesn’t quite capture the essence of it. While this definitely makes understanding Integral calculus easier, Integration, as a tool of mathematics, came before differentiation. Integral calculus dates back to 5^th century BC when the area of a shape was obtained by imagining small polygons forming the shape which was called the method of exhaustion. It is widely debated who proposed the modern theory of integration. While most believe it was Leibniz who introduced a proper theory of integration, many believe Isaac Newton was behind the advancements in this very important tool of mathematics. Integration as a tool was needed mainly to calculate area under a curve.

Integral Calculus Symbol

The symbol ∫ represents integration. The symbol dx, called the differential of the variable x, indicates a very small instant which is the step size of the variable of integration, x. ∫ is taken from a letter which stands for summa or sum or total. This means an integral is a sum of area of infinitesimally small rectangles under the curve, length- f(x) & width-dx, being considered for integration over the variable x.

Fourier was the first person known to use limits on the top & bottom of the integral symbol, to mark the start & the end point of integration. The value on the bottom representing the start & that on the top represents the end point. So, the integral symbol with a & b as the marked limits basically represents area under the curve f(x) between these two values of the x. This form of integral is known as Definite Integral & is the more applied form of Integration.

Definite Integral Calculus

Definite integral represents the real world application of integration, since it is mostly finite area which we are looking to calculate. This is also one of the most asked topics in JEE. One needs an eye to identify the category to which a problem of definite integral belongs. There is one and only one way to develop oneself in any form of integration i.e. practice as many problems as possible. Important formulae of definite integral are same as those of the indefinite integral. It is the properties of definite integrals which play a key role in solving problems efficiently. Here are few of the most important properties of definite Integrals.

integration fundamentals — Properties of Definite Integrals

There are functions which can be integrated over definite limits, using the properties listed above, but their indefinite integrations do not exist.

Indefinite Integral Calculus

When there are no limits marked on the integral symbol, the result of integration is known as indefinite integral & is a generalised result of the area under the curve f(x) for the variable x, wherever it is continuous. If limits are applied to the result of indefinite integration, we obtain its definite value. Indefinite Integral is known to have a wide range of applications as well. Most of the theory in Physics & Chemistry is a result of Indefinite Integration. Let’s have a look at some of the important formulae in indefinite integration.

Indefinite Integration Important Formulae

Integration Formulas

The integration of a function f(x) is given by F(x) and is represented as

∫f(x) dx = F (x) + C

Where, the Right Hand Side of the equation means integral of f(x) with respect to x.

F(x) is called anti-derivative or primitive.

f(x) is called the integrand.

dx is called the integrating agent.

C is an arbitrary constant known as the constant of integration.

x is the variable of integration.

We are aware of the Anti-derivatives of the basic functions. The integrals of these functions can be obtained by us.

Let’s discuss a few important formulae and their applications in determining the integral value of other functions.

integration — How to integrate complex expressions?

Proof: We can express the integrand as:

Multiplying the numerator and the denominator by 2a and simplifying the obtained expression we get:

Hence, upon integrating the obtained expression with respect to x, we get

According to the properties of integration, the integral of sum of two functions is equal to the sum of integrals of the given functions, i.e.

Therefore, equation 1 can be represented as:

Integrating with respect to x we get:

Proof: The integrand can be expressed as:

Multiplying the numerator and the denominator by 2a and simplifying the obtained expression we get:

Manipulating the expression for our convenience

Hence, upon integrating the obtained expression with respect to x we get:

According to the properties of integration, the integral of sum of two functions is equal to the sum integrals of the given functions, i.e.

Therefore, the equation 2 can be rewritten as

Integrating with respect to x, we get:

Proof: Let x = a tan Ɵ. Differentiating both sides of this equation with respect to x we have;

dx = a sec²Ɵ dƟ

Hence, using this, the integral can be expressed as:

Integrating with respect to x we get:

Proof: Let x = a tan Ɵ. Differentiating both sides of this equation with respect to x we have;

dx = a sec²Ɵ dƟ

Therefore, using this, the integral can be expressed as:

Using the trigonometric identity sec²Ɵ = 1 + tan²Ɵ, the above equation can be written as:

Integrating with respect to x, we have:

Substituting the value of Ɵ in the above equation we get;

Here, C = C₁ – log |a|

Proof: Let x = a sin Ɵ. Differentiating both sides of this equation with respect to x we have;

dx = a cosƟ dƟ

Hence, using this, the integral can be expressed as:

Using the trigonometric identity 1 – sin²Ɵ =cos²Ɵ, the above equation can be written as:

Integrating with respect to x, we have:

Substituting the value of Ɵ in the above equation we get;

7. Integration by parts:

Integration by parts is the integration technique most commonly used when you have to integrate the product of two functions.

The generalised formula for integration by parts can be given as:

Let’s understand with an example. There are many a instances when a given function is a product of two simpler functions.

∫x sinx dx is a product of two different simpler functions x. To integrate such product of functions, the integration by parts rule is used. In this example, v= x and du/dx =sin x.

Derivation of formula:

This final formula is known as integration by parts.

Now, let us solve the above example using this formula:

I=x∫sin⁡xdx−∫(ddx(x)∫sin⁡xdx)dx

I=−xcosx−∫(1⋅∫sinxdx)dx

⟹I=−xcos⁡x+∫cos⁡xdx

⟹I=−xcosx+sinx+C

In case the final product is still a product a two integrals, we have to use the above formula again to simplify the integral.

8. Integration by substitution:

Also known as the Reverse Chain Rule, this is commonly used to solve relatively complex integrals.

Here, g’(x) is the derivative of g(x). In case the integral is of the above form, it can be simplified by substituting g(x) as u. hence the function now becomes ∫_a^bf(u)du
The above form is a simple integration, where f(u) can be integrated and u can finally be substituted back as g(x).

Let us understand this with an easy example.

∫sin(a x + b) dx

Let us substitute (ax+b) with u.

So, differentiating this, we get du/dx=a.

Or, dx=1/a du.

Now substituing these above, we get:

I = (1/a) ∫sin(u) du

= (1/a) cos(u)

= (1/a) cos(a x + b) + K

How to Approach Integration

Integration, like any other topic of mathematics, should be practiced as much as possible. One thing special about integration is that there is almost always more than one way of solving a given problem. But, there is only one quickest way to solve it. In a time-bound test, it is very important that one’s solution is as quick as possible. This is the ‘having an eye for integral problems’ talked about earlier. Here are some standard ways to prepare yourself for JEE integral calculus.

Learn to identify the Form

Instead of jumping right away on a problem, it is best to judge the form it is in. In the beginning, this just means you have to stare at the problem and identify the form it is in. This will keep you from going astray in the problem.

Rearrange the Integral

There are times when an integration problem seems difficult to solve but a simple rearrangement can make solving it very convenient. Generally, a seemingly complex problem in JEE can easily be solved by rearrangement. Basic rearrangement involves simple steps which don’t change the problem. Some of the simplest rearrangements are as follows

Adding & Subtracting
Multiply & Dividing
Breaking the Numerator

Memorise the Conventional Formulae

It might sound mildly offensive to some aspirants when asked to ‘memorise‘ the formulae, but memorising a formula cuts a lot on time. You should of course try and derive the formula before memorising it. This will offer you a definite edge over the others.

Know the Standard Integration Techniques

All aspirants must be aware of the standard integration techniques such as Substitution, By Parts, Method of Partial Fractions etc. This can only be achieved by practicing problems from each of these techniques.

Diversified Practice

Other than practicing from NCERT, one should make sure they practice from at least two other different books. This is to ensure exposure to a different type of problems. NCERT is, of course, must because there have been times when JEE asked a direct problem from NCERT Mathematics in integration. ‘Problems in Calculus of One Variable’ by I.A. Maron is one of the highly recommended bookson calculus.

Find out an advanced guide to calculus in this article!

Written by kundariya RISHIT

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