Properties of Definite Integrals
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
There are many properties regarding definite integral. We will discuss each property one by one with proof.
Definite Integrals
Definite integrals are used to compute the area under a curve within a given interval. They have several important properties that simplify calculations and provide deeper insights into the behavior of functions.
Read More: Definite Integrals
Here are some key properties of definite integrals:
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) dxHence, 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)dyFrom 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 = I2Replacing 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) dxThis can be further written as
\int_{0}^{2a} f(x) dx = 2\int_{0}^{a} f(x) dxPart 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) dxThis 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)dyUsing property 2, I1 can be written as
I1 =
\int_{0}^{a} f(-y)dyand using property 1 I1 can be written as
I1 =
\int_{0}^{a} f(-x)dxPutting 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) dxPart 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 =
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 =
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 =
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 =
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)
Related Articles:
- Evaluating Definite Integrals
- Definite Integral | Definition, Formula & How to Calculate
- Definite Integral
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.
Summary
Definite integrals possess several important properties that simplify calculations and analysis. These include linearity, which allows the integration of linear combinations of functions; additivity over intervals, enabling the splitting of integrals into smaller parts; reversal of limits, which changes the sign of the integral when upper and lower bounds are swapped; integration of even and odd functions over symmetric intervals; the comparison property for comparing integrals of functions with known relationships; and the mean value theorem for integrals, which relates the integral to an average function value. These properties provide powerful tools for evaluating integrals, analyzing functions, and solving various mathematical and real-world problems.
