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

Expert Answers
jeew-m eNotes educator| Certified Educator


`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]`

lfryerda eNotes educator| Certified Educator

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


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