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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 `