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Properties of Definite Integrals

Last Updated : 30 Sep, 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+ 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, $\int_{0}^{a}$ f(2a – x)dx can be written as
$-\int_{2a}^{a}$ f(y) dy = I₂
Replacing equation (1) with the value of I₂we 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)

$\text{[math]}$ = 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, I₁can be written as
I₁ = $\int_{0}^{a}$ f(-y)dy
and using property 1 I₁can 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)⁹⁹dx

Solution:

Using property 4.2 he given question can be written as
$\int_{0}^{1}$ (1 – x) [1 – (1 – x)]⁹⁹dx
$\int_{0}^{1}$ (1 – x) [1 – 1 + x]⁹⁹dx
$\int_{0}^{1}$ (1 – x)x⁹⁹dx
$\begin{bmatrix} \frac{x^{100}}{100} - \frac{x^{101}}{101} \end{bmatrix}_{0}^{1}$
= 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