Properties of Definite Integrals (original) (raw)
Last Updated : 14 May, 2026
A definite integral has fixed upper and lower limits and is used to find the area under a curve over a given interval. It also has several useful properties that make calculations easier and help in understanding functions.
\int_{a}^{b}f(x) = F(b) − F(a)
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 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 = 2\int_{0}^{a}f(x) dx;
**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
Solved 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]20
= -[0/2 - 1/2] + [4/2 - 0]
= 1/2 + 2
= 5/2
**Example 5 : Evaluate ∫[-π to π] cos(x)dx and ∫[-π to π] sin(x)dx
**Solution :
Solution:
cos(x) is even:
∫[-π to π] cos(x)dx = 2∫[0 to π] cos(x)dx = 2[sin(x)]|[0 to π] = 2(0 - 0) = 0
sin(x) is odd:
∫[-π to π] sin(x)dx = 0 (by property of odd functions)
Practice Problems
****1).**Evaluate ∫[0 to 4] (2x + 3)dx using the properties of definite integrals.
****2).**If ∫[0 to 2] f(x)dx = 5 and ∫[2 to 4] f(x)dx = 7, find ∫[0 to 4] f(x)dx.
**3).Calculate ∫[-1 to 1] |x|dx using the properties of even functions.
****4).**Given that ∫[0 to 1] f(x)dx = 2, evaluate ∫[0 to 1] [3f(x) - 2]dx.
****5).**Prove that ∫[0 to π/2] sin(x)dx = ∫[0 to π/2] cos(x)dx using properties of definite integrals.
****6).**If ∫[0 to 1] f(x)dx = 3 and ∫[0 to 1] g(x)dx = 2, calculate ∫[0 to 1] [2f(x) - g(x)]dx.
**7).Show that ∫[-a to a] x³dx = 0 for any positive real number a.
**8).Given ∫[0 to 1] f(x)dx = 2 and ∫[1 to 2] f(x)dx = 3, find ∫[2 to 0] f(x)dx.
****9).**Evaluate ∫[0 to π] sin²(x)dx using the identity sin²(x) = (1 - cos(2x))/2 and properties of definite integrals.
**10).If ∫[0 to 1] f(x)dx = 4 and ∫[0 to 2] f(x)dx = 10, calculate ∫[1 to 2] f(x)dx.