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`
Posted on
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