We list below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily.
`color {red} {P_0 : int_a^b f(x) dx = int_a^b f(t) dt}`
`color {red} {P_1 : int_a^b f(x) dx = -int_b^a f(x) dx` . (In particular , `int_a^a f(x) dx = 0)`
`color {red} {P_2 : int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx}`
`color {red} {P_3 : int_a^b f(x) dx = int_a^b f(a +b -x ) dx}`
`color {red} {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`)
`color {red} {P_5 : int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a -x ) dx}`
`color {red} {P_6 : {tt((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)))`
`color {red} {P_7 : {tt ((int_(-a)^a f(x) dx = 2 int_0^a f(x) dx, "if f is an even function"),( 0 , "if f is an odd function"))`
`"Proof:"` of `P_0` It follows directly by making the `color{green}{"substitution x = t."}`
`"Proof"` of `color {red} {P_1}` 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^b f(x) dx = 0`
`color {red}{ "Proof" }` of `P_2` 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`
`"Proof"` of `P_3` Let `t = a + b – x,`
Then `dt = – dx.` When `x = a, t = b` and when `x = b, t = a.`
`int_a^b f(x) dx = - int_b^a f( a +b - t) dt`
`= int_b^a f( a + b - t ) dt`
`= int_a^b f( a + b - x ) dx ` by `P_0`
`"Proof"` of `P_4` Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now proceed as in `P_3`
`"Proof"` of `P_5` 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`
`"Proof"` of `color {red} {P_6}` Using `color {red} {P_5}`, we have
`int_0^(2a) f(x) dx = int_0^(2a) 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) d = 0`
`"Proof"` of `color {red} {P_7}` Using `color {red} {P_2}`, we have
`int_(-a)^a f(x) dx = int_(-a)^0 f(x) dx + int_0^a f(x)`, 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^af(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) + 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) f(x) dx = - int_0^a f(x) dx + int_0^a f(x) dx = 0`
We list below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily.
`color {red} {P_0 : int_a^b f(x) dx = int_a^b f(t) dt}`
`color {red} {P_1 : int_a^b f(x) dx = -int_b^a f(x) dx` . (In particular , `int_a^a f(x) dx = 0)`
`color {red} {P_2 : int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx}`
`color {red} {P_3 : int_a^b f(x) dx = int_a^b f(a +b -x ) dx}`
`color {red} {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`)
`color {red} {P_5 : int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a -x ) dx}`
`color {red} {P_6 : {tt((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)))`
`color {red} {P_7 : {tt ((int_(-a)^a f(x) dx = 2 int_0^a f(x) dx, "if f is an even function"),( 0 , "if f is an odd function"))`
`"Proof:"` of `P_0` It follows directly by making the `color{green}{"substitution x = t."}`
`"Proof"` of `color {red} {P_1}` 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^b f(x) dx = 0`
`color {red}{ "Proof" }` of `P_2` 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`
`"Proof"` of `P_3` Let `t = a + b – x,`
Then `dt = – dx.` When `x = a, t = b` and when `x = b, t = a.`
`int_a^b f(x) dx = - int_b^a f( a +b - t) dt`
`= int_b^a f( a + b - t ) dt`
`= int_a^b f( a + b - x ) dx ` by `P_0`
`"Proof"` of `P_4` Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now proceed as in `P_3`
`"Proof"` of `P_5` 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`
`"Proof"` of `color {red} {P_6}` Using `color {red} {P_5}`, we have
`int_0^(2a) f(x) dx = int_0^(2a) 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) d = 0`
`"Proof"` of `color {red} {P_7}` Using `color {red} {P_2}`, we have
`int_(-a)^a f(x) dx = int_(-a)^0 f(x) dx + int_0^a f(x)`, 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^af(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) + 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) f(x) dx = - int_0^a f(x) dx + int_0^a f(x) dx = 0`