# 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 = I

_{1 }+ I_{2 }(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 = I

_{2}Replacing equation (1) with the value of I

_{2 }we getf(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

I

_{1 }= f(-y)dyUsing property 2, I

_{1 }can be written asI

_{1}= f(-y)dyand using property 1 I

_{1 }can be written asI

_{1 }= f(-x)dxPutting value of I

_{1}in equation (1), we getf(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)x

^{99 }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]

^{2}_{1 }+ 2[x]^{3}_{2 }+ 3[x]^{4}_{3 }+ 4[x]^{5}_{4}= 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]

= -[x

^{2}/2]^{0}_{-1 }+ [x^{2}/2]^{2}_{0 }= -[0/2 – 1/2] + [4/2 – 0]

= 1/2 + 2

= 5/2

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