# Properties of Definite Integrals

Last Updated : 27 Feb, 2024

An integral that has a limit is known as a definite integral. It has an upper limit and a lower limit. It is represented as

f(x) = F(b) âˆ’ F(a)

There are many properties regarding definite integral. We will discuss each property one by one with proof.

## Properties of Definite Integrals

Various properties of the definite integrals are added below,

### Property 1:  f(x) dx =  f(y) dy

Proof:

f(x) dx…….(1)

Suppose x = y

dx = dy

Putting this in equation (1)

f(y) dy

### Property 2: f(x) dx = – f(x) dx

Proof:

f(x) dx = F(b) – F(a)……..(1)

f(x) dx = F(a) – F(b)………. (2)

From (1) and (2)

We can derive  f(x) dx = – f(x) dx

### Property 3:f(x) dx = f(x) dx + f(x) dx

Proof:

f(x) dx = F(b) – F(a) ………..(1)

f(x) dx = F(p) – F(a) ………..(2)

f(x) dx = F(b) – F(p) ………..(3)

From (2) and (3)

f(x) dx +f(x) dx = F(p) – F(a) + F(b) – F(p)

f(x) dx + f(x) dx = F(b) – F(a) = f(x) dx

Hence, it is Proved.

### Property 4.1: f(x) dx = f(a + b – x) dx

Proof:

Suppose

a + b – x = y…………(1)

-dx = dy

From (1) you can see

when x = a

y = a + b – a

y = b

and when x = b

y = a + b – b

y = a

Replacing by these values he integration on right side becomes  f(y)dy

From property 1 and property 2 you can say that

f(x) dx = f(a + b – x) dx

### Property 4.2: If the value of a is given as 0 then property 4.1 can be written as

f(x) dx = f(b – x) dx

### Property 5:  f(x) dx = f(x) dx + f(2a – x) dx

Proof:

We can write  f(x) dx as

f(x) dx = f(x) dx + f(x) dx  ………….. (1)

I = I1 + I

(from property 3)

Suppose 2a – x = y

-dx = dy

Also when x = 0

y = 2a, and when x = a

y = 2a – a = a

So,  f(2a – x)dx  can be written as

f(y) dy = I2

Replacing equation (1) with the value of I2 we get

f(x) dx = f(x) dx + f(2a – x) dx

### =  0; if f(2a – x) = -f(x)

Proof:

From property 5 we can write  f(x) dx as

f(x) dx =f(x) dx + f(2a – x) dx  ………….(1)

Part  1: If f(2a – x) = f(x)

Then equation (1) can be written as

f(x) dx =f(x) dx + f(x) dx

This can be further written as

f(x) dx = 2 f(x) dx

Part  2: If f(2a – x) = -f(x)

Then equation (1) can be written as

f(x) dx=f(x) dx – f(x) dx

This can be further written as

f(x) dx= 0

### = 0; if a function is odd i.e. f(-x) = -f(x)

Proof:

From property 3 we can write

f(x) dx as

f(x) dx = f(x) dx + f(x) dx  ………(1)

Suppose

f(x) dx = I1 ……(2)

Now, assume x = -y

So, dx = -dy

And also when x = -a then

y= -(-a) = a

and when x = 0 then, y = 0

Putting these values in equation (2) we get

I1  f(-y)dy

Using property 2, I1 can be written as

I1 f(-y)dy

and using property 1 I1 can be written  as

I1  f(-x)dx

Putting value of I1 in equation (1), we get

f(x) dx = f(-x) dx +f(x) dx   ……….(3)

Part 1: When f(-x) = f(x)

Then equation(3) becomes

f(x) dx = f(x) dx + f(x) dx

f(x) dx = 2f(x) dx

Part  2: When f(-x) = -f(x)

Then equation 3 becomes

f(x) dx = –f(x) dx +f(x) d

f(x)dx = 0

## Example on Properties of Definite Integrals

Example 1: I =  x(1 – x)99 dx

Solution:

Using  property  4.2 he given question can be written as

(1 – x) [1 – (1 – x)]99 dx

(1 – x) [1 – 1 + x]99 dx

(1 – x)x99 dx

= 1/100 – 1/101

= 1 / 10100

Example 2: I =  cos(x) log

Solution:

f(x) = cos(x) log

f(-x) = cos(-x) log

f(-x) = -cos(x) log

f(-x) = -f(x)

Hence the function is odd. So, Using property

f(x)dx = 0; if a function is odd i.e. f(-x) = -f(x)

cos(x) log  = 0

Example 3: I =  [x] dx

Solution:

0 dx + 1 dx +  2 dx + 3 dx +  4 dx  [using Property 3]

= 0 + [x]21 + 2[x]32  + 3[x]43 + 4[x]54

= 0 + (2 – 1) + 2(3 – 2) + 3(4 – 3) + 4(5 – 4)

= 0 + 1 + 2 + 3 + 4

= 10

Example 4: I =  |x| dx

Solution:

(-x) dx +  (x) dx  [using Property 3]

= -[x2/2]0-1 + [x2/2]2

= -[0/2 – 1/2] + [4/2 – 0]

= 1/2 + 2

= 5/2

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