Properties of Definite Integral
Property 1:[Tex] \bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(z)dz}[/Tex]
Proof:
Let p(x) be a antiderivative of f(x). Then,
[Tex]\frac{d}{dx} [/Tex]{p(x)} = f(x) ⇒ \[Tex]\frac{d}{dz} [/Tex]{p(z)} = f(z)
[Tex]\int\limits_{a}^b f(x)dx = \big[p(x)\big]_a^b [/Tex] = p(b) – p(a) ——————- (i)
and [Tex]\int\limits_{a}^b f(z)dz = \big[p(z)\big]_a^b [/Tex] = p(b) – p(a) ——————-(ii)
From (i) and (ii)
[Tex]\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(z)dz}[/Tex]
Property 2: [Tex]\bold{\int\limits_{a}^{b}f(x)dx =-\int\limits_{b}^{a}f(x)dx} [/Tex]
If the limits of the definite integral are interchanged then, its value changes by a minus sign only.
Proof:
Let p(x) be the antiderivative of f(x). Then,
[Tex]\int\limits_{a}^b f(x)dx [/Tex] = p(b) – p(a)
and [Tex]-\int\limits_{b}^a f(x)dx [/Tex] = -[p(a) – p(b)] = p(b) – p(a)
[Tex]\bold{\int\limits_{a}^{b}f(x)dx =-\int\limits_{b}^{a}f(x)dx} [/Tex]
Property 3:[Tex] \bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx} [/Tex]where a < c < b
Proof:
Let p(x) be the antiderivative of f(x). Then,
[Tex]\int\limits_{a}^b f(x)dx [/Tex] = p(b) – p(a) ——————(i)
[Tex]\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx [/Tex] = [p(c) – p(a)] + [p(b) – p(c)] = p(b) – p(a) ——————(ii)
From (i) and (ii)
[Tex]\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx} [/Tex]
Property 4: [Tex]\bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}[/Tex]
Proof:
Let x = a – t . Then, dx = d(a – t) ⇒ dx = -dt
When x = 0 ⇒ t = a and x = a ⇒ t = 0
[Tex]\int\limits_{0}^{a}f(x)dx =-\int\limits_{a}^{0}f(a – t)dt[/Tex]
⇒ [Tex]\int\limits_{0}^{a}f(x)dx =\int\limits_{0}^{a}f(a – t)dt [/Tex] [ By second property ]
⇒ [Tex]\int\limits_{0}^{a}f(x)dx =\int\limits_{0}^{a}f(a – x)dx [/Tex] [ By first property ]
[Tex]\bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}[/Tex]
Property 5: [Tex]\bold{\int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}[/Tex]
Proof:
Using third property
[Tex]\int\limits_{-a}^{a}f(x)dx=\int\limits_{-a}^{0}f(x)dx+\int\limits_{0}^{a}f(x)dx [/Tex] ——————–(i)
Let x = – t , dx = -dt
Limits : x= -a ⇒ t = a and x = 0 ⇒ t = 0
[Tex]\int\limits_{-a}^{0}f(x)dx=\int\limits_{a}^{0}f(-t)(-dt) =-\int\limits_{a}^{0}f(-t)dt=\int\limits_{0}^{a}f(-t)dt [/Tex] [By second property]
⇒[Tex] \int\limits_{-a}^{0}f(x)dx=\int\limits_{0}^{a}f(-x)dx [/Tex] [By first property] ———–(ii)
From (i) and (ii)
[Tex]\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}f(-x)dx+\int\limits_{0}^{a}f(x)dx[/Tex]
⇒[Tex]\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}[f(-x)+f(x)]dx[/Tex]
⇒ [Tex]\int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(-x) = f(x) \\ 0 & , if f(-x) = -f(x) \end{cases}[/Tex]
⇒ [Tex]\bold{\int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}[/Tex]
Property 6: If f(x) is a continuous function defined on [0, 2a],
[Tex]\bold{\int\limits_{0}^{2a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a – x) = f(x) \\ 0 & , if f(2a – x) = -f(x)\end{cases}}[/Tex]
Proof:
Using third property
[Tex]\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{a}^{2a}f(x)dx [/Tex] —————–(i)
Consider [Tex]\int\limits_{a}^{2a}f(x)dx [/Tex]
Let x = 2a – t , dx = -d(2a – t) ⇒ dx = -dt
Limits : x= a ⇒ t = a and x = 2a ⇒ t = 0
[Tex] \int\limits_{a}^{2a}f(x)dx=-\int\limits_{a}^{0}f(2a – t)dt [/Tex]
⇒ [Tex] \int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a – t)dt [/Tex] [ Using second property]
⇒ [Tex] \int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a – x)dx [/Tex] [ Using first property]
Substituting [Tex] \int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a – x)dx [/Tex]in (i)
[Tex]\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{0}^{a}f(2a – x)dx = \int\limits_{0}^{a}[f(x) + f(2a – x)]dx[/Tex]
[Tex]\bold{\int\limits_{0}^{2a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a – x) = f(x) \\ 0 & , if f(2a – x) = -f(x)\end{cases}}[/Tex]
Property 7: [Tex]\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b – x)dx}[/Tex]
Proof
Let t = a + b – x ⇒ dt = -dx
Limits : x = a , y = b and x = b , y = a
After putting value and limit of t in [Tex]\int\limits_{a}^{b}f(a + b – x)dx[/Tex]
⇒[Tex]\int\limits_{a}^{b}f(a + b – x)dx =-\int\limits_{b}^{a}f(t)dt [/Tex]
⇒ [Tex]\int\limits_{a}^{b}f(a + b – x)dx =\int\limits_{a}^{b}f(t)dt [/Tex] [Using second property]
⇒ [Tex]\int\limits_{a}^{b}f(a + b – x)dx =\int\limits_{a}^{b}f(x)dx [/Tex] [Using first property]
[Tex]\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b – x)dx}[/Tex]
Also, Check
How to Calculate a Definite Integral?
A definite integral is a mathematical concept used in calculus to calculate the total effect of a function over a given time frame. It represents the net area between the function’s graph and the x-axis during a certain time frame. A definite integral is calculated by determining the area under a curve between two specified locations on the x-axis. The definite integral is the net accumulation of the function’s values over the specified interval. This article talks about how to calculate definite integral and also provide some solved examples based on calculation of definite integration.