Properties of Definite Integrals

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

$\int_{a}^{b}$ f(x) = F(b) − F(a)

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

Properties of Definite Integrals

Various properties of the definite integrals are added below,

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 = 2 $\int_{0}^{a}$ 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, 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

Example on Properties of Definite Integrals

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