Try the substitution u= 1- x. Then becomes which is easy to integrate. But you also have to replace "dx". With u= 1- x, du/dx= -1 so du= -dx and dx= -du. Your integral becomes .

Notice that I didn't put the limits of integration here. You could integrate , to get , of course, change that to and evaluate that at x=0 and x=1.

But I prefer to change the limits of integration as I substitute. Those are x= 0 and x= 1 and they have to be changed to u also. When x= 0, u= 1 and when x= 1, u= 0.

That "-" in "1- x" has reversed the order of the limits of integration- but so you really have

and the negative has disappeared!

Another way of looking at it is to remember that integration is the opposite of differentiation. Whatever function you integrated to, you would have f(u)= f(1- x). And you would use the "chain rule" to differentiate: the derivative of f(1-x) is the derivative of f times the derivative of 1- x, which is -1.

With any substitution for adecreasingfunction, like 1- x, the derivative will be negative. That's why you need the "-" in the anti-derivative.