Properties of Definite Integrals

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

f(x) = F(b) − F(a)

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

Properties

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 + I₂ 

(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

Property 6 :  f(x) dx = 2f(x) dx; if f(2a – x) = f(x)

                                                      =  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

Property 7:  f(x) dx =f(x) dx; if a function is even i.e. f(-x) = f(x)

                                                = 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

Examples

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]²₀ 

= -[0/2 – 1/2] + [4/2 – 0]

= 1/2 + 2

= 5/2