# If g(x) is continuous on [0,1], show that integrate from 0 to 1 of g(1-x)dx = integrate from 0 to 1 of g(x)dx

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Since `g(x)` is continuous on the interval `[0,1]` , it means that we can use substitution on the interval. This means that

`int_0^1 g(1-x)dx` let `u=1-x` then `du=-dx` and at `x=0`

`u=1` and at `x=1` , `u=0` so the integral becomes

`=-int_1^0g(u)du` switch the limits, which changes the sign

`=int_0^1 g(u)du` but, the integral variable is arbitrary, so replace u with x

`=int_0^1g(x)dx`

**This means the integral `int_0^1g(1-x)dx=int_0^1g(x)dx` as requested.**

**Sources:**

let;

`1-x = t`

`-1 = dt/dx`

` dx = -dt`

When x=0 then t=1

When t=1 then x = 0

`int_0^1g(1-x)dx `

`= int_1^0 g(t)(-1)dt`

`= -int_1^0 g(t)dt`

Now lets replace t by x.

`-int_1^0 g(t)dt = = -int_1^0 g(x)dx`

`-int_1^0 g(x)dx `

`= -[P(x)]_1^0` Here `(dP(x))/dx = g(x)`

`= -[P(0)-P(1)]`

`= P(1)-P(0)`

`-int_1^0 g(x)dx = P(1)-P(0)` ------(1)

`int_0^1 g(x)dx`

`= [P(x)]_0^1`

`= P(1)-P(0) `

`int_0^1 g(x)dx = P(1)-P(0)` ----(2)

(1) = (2)

`int_0^1 g(x)dx = -int_1^0 g(t)dt`

`int_0^1 g(x)dx = int_0^1g(1-x)dx `

This happens only since g(x) is continuous over `x in [0,1]`

**Sources:**