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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » How to Measure Definite Integral

How to Measure Definite Integral

Study material notes on how to measure definite integral, the definition of modulus of definite integral, properties of modulus of definite integral and other related topics in detail.

Table of Content
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Integration is the sum of the area and the definite integral is used as a tool to identify the area within an interval or limit. The mathematical expression of definite integral is ∫baf(x)dx.

This article covers the measurement of definite integral, the concept of definite integral, a thorough explanation on properties of definite integral, formula of definite integral and so on.   

Definition of Definite Integral

A definite integral is given using this notation here, where is a function f (x) is continuous on the closed interval, from a to b, [ a, b], then the definite integral from a to b is defined as below:

 ∫ab  f (x) dx = limn→∞ ∑nr=1 f (ci) ΔXi

Properties of Definite Integral

In this section you will find a brief of the basic properties of definite integral, its functions to understand the definite integral more effectively. So, let’s have a walk through these properties.

1.  ∫ba f(x) dx= ∫ca f(x) dx + ∫bc f(x) dx

Proof:  Let g (x) be the antiderivative of f(x). Then,

LHS = g(b) – g(a)

RHS = (g (c) – g (a) ) + (g (b) – g (c) )

     = g(b) – g(a)

2.  ∫ba f(x) dx = -∫abf(x) dx also, ∫aa f(x) dx = 0

Proof: Let g (x) be the antiderivative of f(x). Then,

LHS = g(b) – g(a)

RHS = – (g (a) – g (b) )

     = g(b) – g(a)

3.  ∫ba f(x) dx = ∫ba f(t) dt

Proof: Let g be the antiderivative of (f). Then,

LHS = g(b) – g(a)

RHS = g(b) – g(a)

Hence proved LHS = RHS

The value of definite integral is always free from the variable of integration, due to this fact we can call the variable of definite integral in definite integral- dummy variable.

4.  ∫ba f(x) dx = ∫ba f (a + b -x) dx

Proof: Take RHS and then, put a+b-x = t, -dx = dt

x- a to b

t- b to a

RHS =  -∫ab f(t) dt = ∫ba f(t) dt =  ∫ba f(x) dx = LHS

5. ∫a0 f(x) dx = ∫a0 f (a-x) dx

Proof : As we can see that property 5 is similar to property 4 we need to put a = 0  and b = a.

This is the most commonly used property in definite integral.

6. ∫2a0 f(x) dx = ∫a0 f (x) dx + ∫a0 f (2a – x) dx

Proof: Using P -I  we will get ….

∫2a0 f(x) dx = ∫a0 f (x) dx + ∫2aa f (x) dx

In I, put x = 2a – t

. dx = – dt

X – a to 2a

T – 2a to 0

∫2a0 f(x) dx = ∫a0 f (x) dx + ∫0a f (2a – t) dt

∫2a0 f(x) dx = ∫a0 f (x) dx + ∫a0 f (2a – x) dx

Specific cases of this property……

a. ∫2a0 f(x) dx = 2 ∫a0 f (x) dx [ if  f (x)  = f (2a – x) ]

b. ∫2a0 f(x) dx = 0  [ if f (x) = -f (2a-x)]

How to calculate a definite integral

Step 1. Calculate the antiderivative F(x)

Step 2. Find values of F(b) and F(a)

Step 3. Calculate F(b) – F(a).

Let’s look at some examples for our better understanding of measurement or calculation of definite integral.

Example 1.  F = ∫64 3x2 dx

= 3∫64 x2 dx  (as per the rule , of a constant getting multiplied by a variable it can be taken outside)

= 3 [x3 / 3]64 ….(as per the power rule ∫ xn dx = x n+1 / (n +1) + c, n ≠ -1)

= [ x3 / ]64

= [6]3 – [4]3  …….( replace x with upper limit and lower limit)

= 216 – 64                           

= 152 

Example 2.  F = ∫42 6x2 – 3x +11 dx

= [ 6x3/3 – 3x2/2 + 11x]4 2 ( 1 dx = x , so 11dx = 11x )

= [ 2x3 – 3/2 x2 + 11x ] 4 2

= [ 2 (4)3 – 3/2 (4)2 + 11 (4)] – [ 2 (2)3 – 3/2 (2)2 + 11 (2)]

= [128 – 24 + 44] – [ 16 – 6 + 22]

=148 – 32

= 116

Conclusion

In this article describing the definite integral, we studied the measurement of the definite integral in much detail. We covered several other topics such as properties of definite integral, important formulas of definite integral and other related topics along with the solved examples. We hope you find this study material helpful for better understanding on measurement of definite integral.

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

What is the use of definite integral?

Ans : For determination of the area under, over, or between curves, de...Read full

What are the rules of definite integrals?

Ans :The sum of the integrals equals to the integral sum, and Difference of th...Read full

Find the value of ∫23 x2 dx.

Ans : Let I = ∫2...Read full

Solve the equation ∫26 4dx

Ans : ∫2...Read full

Solve the equation, given that ∫03 x2 dx = 8, find ∫03 -4x2 dx?

Ans : ∫03...Read full

Ans : For determination of the area under, over, or between curves, definite integral can be used. If a function is positive, the area between it and the x axis is simply the definite integral. If the function is negative, then the area is -1 times of the definite integral.

Ans :The sum of the integrals equals to the integral sum, and Difference of the integrals equals to the integrals difference, and

The integral of the product of a constant and function is the constant multiplied by the integral of the function.

  

Ans : Let I = ∫23 x2 dx

Now, ∫x2 dx = (x3)/3

Now, I =  ∫23 x2 dx = [(x3)/3]23

= (33)/3 – (23)/3

= (27/3) – (8/3)

= (27 – 8)/3

= 19/3

Therefore, ∫23 x2 dx = 19/3      

Ans : ∫26 3dx = 4 (6-2)

                     = 16

Ans : ∫03 -4x2 dx = -4 ∫03 x2 dx

                                 = -4 .8

                                 = – 32

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