Let ;

`intf(x)dx = g(x)`

Then;

`int_a^b f(x)dx `

`= [g(x)]^b_a`

`= g(b)-g(a)`

`int_a^b f(x)dx = = g(b)-g(a) -----(1)`

let;

`t = a+b-x`

`(dt)/dx = -1`

` -dt = dx`

When x = a then t = b

When x = b then t = a

`int^b_a f(a+b-x)dx`

`= int^a_b f(t)(-dt)`

`= -int^a_b f(t)dt` This is same as `int_a^b f(x)dx` .

`= -[g(t)]^a_b`

`= -[g(a)-g(b)]`

`= g(b)-g(a)`

`int^b_a f(a+b-x)dx = g(b)-g(a) -----(2)`

(1) = (2)

*Therefore;*

`int_a^b f(x)dx = int^b_a f(a+b-x)dx`

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