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Properties of Definite Integrals

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1) `P_0 : int_a^b f(x) dx = int_a^b f(t) dt`
Proof: It follows directly by making the substitution x = t. 

2) `P_1: int_a^b f(x) dx =- int_b^a f(x) dx. "In particular" , int_a^a f(x) dx = 0`
Proof:  Let F be anti derivative of f. Then, by the second fundamental theorem of
calculus, we have `int_a^b f(x) dx = F(b) - F(a) = [F(a)-F(b)] = -int_b^a f(x) dx`
Here, we observe that, if a = b, then `int_a^a f(x) dx = 0`

3) `P_2 : int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx` 
Proof:  Let F be anti derivative of  f. Then
`int_a^b f(x) dx = F(b) - F(a)`         ...(1)
`int_a^c f(x) dx  = F(c) - F(a)`         ...(2) and 
`int_c^b f(x) dx = F(b) - F(c)`         ...(3) 
Adding (2) and (3), we get 
`int_a^c f(x) dx + int_c^b f(x) dx = F(b) - F(a) = int_a^b f(x) dx `
This proves the property `P_2.`

4) `P_3 : int_a^b f(x) dx = int_a^b f(a + b - x) dx `
Proof:  Let t = a + b – x. Then dt = – dx. When x = a, t = b and when x = b, t = a. Therefore
`int_a^b f(x) dx = - int _b^a f(a+b-t) dt`
= `int_a^b f(a+b-t)dt    ("by"  P_1)`
=`int_a^b f(a+b-x) dx     ("by"  P_0)`

5) `P_4: int_0^a f(x) dx = int_0^a f(a - x) dx `
(Note that `P_4` is a particular case of `P_3`)
Proof:  Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now proceed as in `P_3`.

6) `P_5: int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a - x) dx `
Proof:  Using `P_2`, we have `int_0^(2a) f(x) dx = int_0^a f(x) dx + int_a^(2a) f(x) dx` 
Let  t = 2a – x in the second integral on the right hand side. 
Then dt = – dx. When x = a, t = a and when x = 2a, t = 0. 
Also x = 2a – t. Therefore, the second integral becomes   
`int _a^(2a) f(x) dx = -int_a^0 f(2a - t) dt  = int_0^a f(2a - t) dt  = int_0^a f(2a - x) dx `

Hence `int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a - x) dx  `

 

7) `P_6: int_0^(2a) f(x) dx = 2 int_0^a f(x) dx , if f(2a - x) = f(x) and 0 if f(2a - x) = -f(x) `
Proof: Using `P_5` , we have `int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a - x) dx ` ...(1)
Now, if    f(2a – x) = f(x), then (1) becomes
`int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(x) dx = 2 int_0^a f(x) dx,`
and if  f(2a – x) = – f(x), then (1) becomes
`int_0^(2a) f(x) dx = int_0^a f(x) dx  - int_0^a f(x) dx = 0 ` 

 

8) `P_7`: 
i)  `int_(-a)^a f(x) dx  = 2 ``int_0^a f(x) dx` if f is an even function , i.e., if f(-x) = f(x)
ii) `int_(-a)^a f(x) dx = 0` , if f is an odd function , i.e., if f(-x) = -f(x).

Proof :
 Using `P_2` we hane 
`int_(-a)^a f(x) dx`  = `int_(-a)^0 f(x) dx` + `int_0^a  f(x) dx`   Then 
t = – x in the first integral on the right hand side. 
dt = – dx. When x = – a, t = a and when 
x = 0, t = 0. Also x = – t.  
Therefore  `int_a^(-a) f(x) dx = - int_(a)^0 f(-t) dt  + int_0^a f(x) dx `
`= int_(0)^a f(-x) dx + int_0^a f(x) dx  `   `("by" P_0)` ...(1) 
(i) Now, if f is an even function, then f(–x) = f(x) and so (1) becomes
` int_(-a)^a f(x) dx  = int_0^a f(x) dx + int_0^a f(x) dx + int_0^a f(x) dx = 2 int_0^a f(x) dx `
(ii) If f is an odd function, then f(–x) = – f(x) and so (1) becomes
`int_(-a)^a f(x) dx = - int_0^a f(x) dx  + int_0^a f(x) dx = 0 `

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