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Properties of Definite Integrals

  • Last Updated : 10 Feb, 2022

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

\int_{a}^{b}   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: \int_{a}^{b}    f(x) dx = \int_{a}^{b}    f(y) dy

Proof:

\int_{a}^{b}    f(x) dx…….(1)

Suppose x = y

           dx = dy

Putting this in equation (1)

\int_{a}^{b}    f(y) dy

Property 2: \int_{a}^{b}    f(x) dx = –\int_{b}^{a}    f(x) dx

Proof:

\int_{a}^{b}    f(x) dx = F(b) – F(a)……..(1)

\int_{b}^{a}    f(x) dx = F(a) – F(b)………. (2)

From (1) and (2) 

We can derive \int_{a}^{b}    f(x) dx = –\int_{b}^{a}    f(x) dx

Property 3: \int_{a}^{b}   f(x) dx = \int_{a}^{p}   f(x) dx + \int_{p}^{b}   f(x) dx

Proof:

\int_{a}^{b}    f(x) dx = F(b) – F(a) ………..(1)

\int_{a}^{p}    f(x) dx = F(p) – F(a) ………..(2)

\int_{p}^{b}    f(x) dx = F(b) – F(p) ………..(3)

From (2) and (3)

\int_{a}^{p}   f(x) dx + \int_{p}^{b}   f(x) dx = F(p) – F(a) + F(b) – F(p)

\int_{a}^{p}   f(x) dx + \int_{p}^{b}   f(x) dx = F(b) – F(a) = \int_{a}^{b}   f(x) dx    

Hence, it is Proved.

Property 4.1: \int_{a}^{b}    f(x) dx = \int_{a}^{b}   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 -\int_{b}^{a}    f(y)dy

From property 1 and property 2 you can say that 

\int_{a}^{b}    f(x) dx = \int_{a}^{b}   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

 \int_{0}^{b}    f(x) dx = \int_{0}^{b}   f(b – x) dx

Property 5: \int_{0}^{2a}    f(x) dx = \int_{0}^{a}   f(x) dx + \int_{0}^{a}   f(2a – x) dx

Proof:

We can write \int_{0}^{2a}    f(x) dx as

\int_{0}^{2a}    f(x) dx = \int_{0}^{a}   f(x) dx + \int_{a}^{2a}   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, \int_{0}^{a}    f(2a – x)dx  can be written as 

-\int_{2a}^{a}    f(y) dy = I2

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

\int_{0}^{2a}    f(x) dx = \int_{0}^{a}   f(x) dx + \int_{0}^{a}   f(2a – x) dx

Property 6 : \int_{0}^{2a}    f(x) dx = 2\int_{0}^{a}   f(x) dx; if f(2a – x) = f(x)

                                                      =  0                    ; if f(2a – x) = -f(x) 

Proof: 

From property 5 we can write \int_{0}^{2a}    f(x) dx as

\int_{0}^{2a}    f(x) dx =\int_{0}^{a}   f(x) dx + \int_{0}^{a}   f(2a – x) dx  ………….(1)

Part  1: If f(2a – x) = f(x) 

Then equation (1) can be written as 

\int_{0}^{2a}   f(x) dx =\int_{0}^{a}   f(x) dx + \int_{0}^{a}   f(x) dx

This can be further written as 

\int_{0}^{2a}    f(x) dx = 2 \int_{0}^{a}   f(x) dx

Part  2: If f(2a – x) = -f(x)

Then equation (1) can be written as 

 \int_{0}^{2a}    f(x) dx= \int_{0}^{a}   f(x) dx – \int_{0}^{a}   f(x) dx

This can be further written as 

\int_{0}^{2a}    f(x) dx= 0

Property 7: \int_{-a}^{a}    f(x) dx =\int_{0}^{a}   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 

\int_{-a}^{a}    f(x) dx as 

\int_{-a}^{a}    f(x) dx = \int_{-a}^{0}   f(x) dx + \int_{0}^{a}   f(x) dx  ………(1)

Suppose 

\int_{-a}^{0}    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 -\int_{a}^{0}    f(-y)dy

Using property 2, I1 can be written as 

I1\int_{0}^{a}    f(-y)dy

and using property 1 I1 can be written  as 

I1 \int_{0}^{a}    f(-x)dx

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

\int_{-a}^{a}    f(x) dx = \int_{0}^{a}   f(-x) dx +\int_{0}^{a}   f(x) dx   ……….(3)

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

Then equation(3) becomes

\int_{-a}^{a}    f(x) dx = \int_{0}^{a}   f(x) dx + \int_{0}^{a}   f(x) dx

\int_{-a}^{a}    f(x) dx = 2\int_{0}^{a}   f(x) dx 

Part  2: When f(-x) = -f(x)

Then equation 3 becomes

\int_{-a}^{a}    f(x) dx = –\int_{0}^{a}   f(x) dx +\int_{0}^{a}   f(x) d

\int_{-a}^{a}    f(x)dx = 0

Examples

Example 1: I = \int_{0}^{1}    x(1 – x)99 dx

Solution:

Using  property  4.2 he given question can be written as 

\int_{0}^{1}    (1 – x) [1 – (1 – x)]99 dx

\int_{0}^{1}    (1 – x) [1 – 1 + x]99 dx

 \int_{0}^{1}   (1 – x)x99 dx

\begin{bmatrix} \frac{x^{100}}{100} - \frac{x^{101}}{101} \end{bmatrix}_{0}^{1}

= 1/100 – 1/101

= 1 / 10100

Example 2: I = \int_{\frac{-1}{2}}^{\frac{-1}{2}}    cos(x) log \begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}

Solution:

f(x) = cos(x) log \begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}

 f(-x) = cos(-x) log \begin{vmatrix} \frac{1-x}{1+x} \end{vmatrix}

 f(-x) = -cos(x) log \begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}

f(-x) = -f(x)

Hence the function is odd. So, Using property 

\int_{-a}^{a}    f(x)dx = 0; if a function is odd i.e. f(-x) = -f(x) 

\int_{\frac{-1}{2}}^{\frac{-1}{2}}    cos(x) log \begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}    = 0

Example 3: I = \int_{0}^{5}    [x] dx

Solution:

\int_{0}^{1}    0 dx + \int_{1}^{2}   1 dx + \int_{2}^{3}    2 dx +\int_{3}^{4}    3 dx + \int_{4}^{5}    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 =  \int_{-1}^{2}    |x| dx

Solution:

\int_{-1}^{0}    (-x) dx + \int_{0}^{2}    (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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