# Mathematics and Integral Definition

Mathematics is a vast field with many branches like geometry, algebra, and calculus. Calculus deals with the change in a quantity or function with respect to some defined or independent variable.

Isaac Newton and Gottfried Leibniz developed modern calculus in the 17th century. Today, calculus is a highly developed and sophisticated field of mathematics that finds its uses in many fields of science. Calculus is used widely in science, economics as well as in engineering.

Here, we will discuss the integral definition and its basic concepts of integration. Then, we will move on to different types of integration. After that, we will learn the various techniques to integrate a function.

## What is Integration?

Integration can be given as follows:

• Integration is the opposite of differentiation, therefore, it is also called antiderivative. Using differentiation, you can find out the gradient of the curve through its equation. However, if you know the gradient of the curve you can find the integral of the function to determine the equation of the curve. Both concepts are highly linked.
• Another way to learn the integral definition is by knowing its basis. Integration, like differentiation, uses the theory of convergence applied to infinite series and sequences.
• The basis of integral definition is that integration essentially divides a curve, function, or a region into infinitely small units. Then the area of each of these units is summed up. The result is called the integral of a function, region, or a curve.

### Notation

Now that we have covered integration definition let’s see how to represent an integral. We write that if:

• dydx=x–(1)

Then,

• y=x dx–(2)

Here, ∫ represents integration.

If you see it written as something like this:

• y=abxdx

Then, a and b represent the limits of integration. This we learn in definite integrals.

In modern calculus, D represents the differential operator whereas 1/D represents the integral operator. So, the equations 1 and 2 can also be written as:

• Dy=x
• y=1Dx

### Uses of Integration

Integration is employed to determine:

• The equation of the curve
• The solution of differential equations
• The area under the curve
• The area between two curves
• The volume of an object
• The centroid of an object
• The electric potential between two charges
• The average value of a curve
• The force applied by liquid pressure, etc.

Due to the above uses, integration is widely used in many branches of engineering and commerce.

### Types of integration

Integration is of two types:

• Indefinite integration
• Definite integration

### Indefinite Integration

The indefinite integral definition is that in such an integral there are no limits. Mathematically:

• y=f(x) dx

The solution of such an integral comes out to be:

• y=F(x)+c

Where “c” is the arbitrary constant, which will always appear at the end of the solution of an indefinite integration.

### Definite Integration

The definite integral definition is that in this case we integrate within a set limit. In other words, the function is integrated within a definite boundary. Mathematically:

• y=abf(x)dx

Its solution will then be of the form:

• y=F(b)-F(a)

The result is a number or a definite quantity with no arbitrary constant in it.

### Integration Techniques: The General Rule

The most basic form of integration is shown in the following equation:

• y=axndx=axn+1n+1+c
•

Where n can never be equal to -1.

Notice that whatever is the power of x, you increase the power of x by 1. At the same time, divide it with the same. This is the mathematical integral definition.

For example, if:

• dydx=3×3

Then:

• y=3x3dx
• y=3×3+13+1+c
• y=3×44+c

A quick note: Here, dy/dx is actually the gradient of a curve and the solution obtained after taking its integral is the equation of the same curve.

Example 2:

• dydx=x
• y=xdx
• y=x3232+c
• y=23×32+c

Example 3:

• dydx=1×3
• y=1x3dx
• y=x-3dx
• y=x-2-2+c
• y=-2×2+c

### Principle of Linearity

You must have come across the principle of linearity in differentiation. It states that when the differential operator is applied on two functions that are added together, the operator can be applied on the individual functions to obtain the solution. Mathematically:

• ddx(axn+bxm)=ddxaxn+ddxbxm

The same rule can be applied in integration as per the integral definition. Mathematically:

• ((axn+bxm)=axn+1n+1+bxm+1m+1+c

For example:

• dydx=x(x+3)
• dydx=x2+3x
• y=x2+3x dx
• y=x2dx+3xdx
• y=x33+3×22+c

### Integration of a Constant

So far, we have learnt different ways to integrate functions having variables and their coefficients. Now, let’s see how to integrate a constant number. We all know that a constant number, say 4, can be written as follows:

• 4=4×0

where x raised to the power 0 equals 1.

Now, if:

• dydx=4
• y=4dx
• y=4x0dx
• y=4×11+c
• y=4x+c

Finding the Value of the Arbitrary Constant

Here’s how to find the value of the arbitrary constant.

If the gradient of a curve is given by dydx=x2+2 and it passes through a point (2,1). Find the equation of the curve:

• dydx=x2+2
• y=x2+2dx
• y=x33+2x+c

This is the equation of the curve. There is another information given in the question; the curve passes through a point (2,1). We will use this information to find out the value of the unknown arbitrary constant.

Since the point (2,1) lies on the curve, it must satisfy its equation. Therefore:

• 1=233+2*3+c
• 1=83+6+c
• c=1-83-6
• c= -233

Now, put this value of c in the equation of the curve to get:

• y=x33+2x-233

### Integrating Whole Powers

In differentiation:

• ddx(ax+b)n+1a(n+1)+c=(ax+b)n

Since, integration is the reverse of differentiation, we can write:

• (ax+b)ndx=(ax+b)n+1a(n+1)+c

For example:

• dydx=(3x+2)5
• y=(3x+2)5dx
• y=(3x+2)63*6+c
• y=(3x+2)618+c

Example 2:

• dydx=(3×2+2x+3)3
• y=(3×2+2x+3)3dx

In this example, you will need to start thinking from the basics. First, apply the general rule of integration i.e. increase the power by one and divide with the same. Then, apply principle of linearity on the function within the brackets. You will get:

• y=(3×2+2x+3)44{3×33+2×22+3x}+c

Simplify the above equation to get the final answer.

### Integration of Trigonometric Functions

Here’s how to calculate the integral of trigonometric functions:

• sinx=-cosx+c
• cosx=sinx+c
• sec2 = tanx+c

### Integration of Exponential Functions

For exponential functions, use the following formulas:

• ex=ex+c
• e-x=e-x+c
• eax+b=1aeax+b+c

### Definite Integral Definition

After mastering the indefinite integration, you can easily find the definite integration. All of the rules that we discussed above will remain the same. The only difference is that through indefinite integration you can determine the equation of a curve or any function. However, with definite integration you can determine numeric values. These values can represent the volume, area, force, speed, velocity, acceleration, etc.

The general equation of definite integration is:

• ddx=[F(x)]=f(x)=abf(x)dx=F(b)-F(a)

Where, abf(x)dx is known as the definite integral.

For example:

• dydx=6x+2

For the interval (2,3):

• y=23(6x+2)dx
• y=[3×2+2x]23
• y=(3*32+2*3)-(3*22+2*2)
• y=33-16
• y=17

### Applications of Integration: Area Under the Curve and Above the Axis

If a curve lies above the x-axis, and its equation is given by f(x)=y. Then, the area of such a curve above the x-axis and between x=a to x=b is given by:

• area= abydx=abf(x)dx

Where f(x) must not be 0 and a ≤ x ≤ b.

For example, the equation of a curve is given by (x)=4×2. Find the area of the curve between the interval 1 ≤ x ≤ 2:

• area= abf(x)dx
• area= 124x2dx
• area=4[-1x]12
• area=4 [(-12)-(-11)]
• area=2 unit2

The same method can be used to determine the area between a curve and the y-axis.

### Calculating the Velocity and Acceleration

In earlier grades, we have studied the relation between displacement speed and velocity. We say that:

• Velocity is the rate of change of displacement
• Acceleration is the rate of change of velocity

Mathematically:

• v=dsdt
• a=dvdt

Where v = velocity of an object, a =acceleration of an object, s= displacement of an object, and t=time in seconds.

So, you can find the velocity of an object by finding the gradient of displacement time graph. Moreover, you can determine the acceleration of an object by finding the gradient of the velocity time graph.

As integration is the reverse of differentiation, you can use the same relations to determine the velocity and acceleration of an object. So, if you need to find the velocity of an object, just integrate the acceleration. Similarly, you can find out the displacement of an object by integrating the velocity.

Mathematically:

• s=vdt

For example, the acceleration of an object is given by a=3-t m/s2. Determine its velocity and displacement at t=2 seconds if at t=0 it moves with a velocity of 8m/s.

• v=(3-t)dt
• v=3t-t22+c

At t=0, v=8m/s

• 8=3*0-022+c
• c=8

Now, at t=2 sec

• v=3*2-222+8
• v=6-2+8=12ms

We can use the equation of velocity to determine the displacement of the object:

• v=3t-t22+c

Where c=8:

• v=3t-t22+8

Now,

• s=vdt
• s=(3t-t22+8)dt
• s=3t2-t36+8t

At t=2 sec

• s=3*22-236+8*2
• s=26.667m

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