If its integral (over the range [0,1]) exists, let it be I, StatusX's hint was that 1-f(x) is the same function but going downhill not uphill, so the integral of 1-f(x) over [0,1] is also I. You can do the rest from here surely. The point is the function is very symmetric, so you can exploit that symmetry, just like you can integrate sin(x) from -t to t for any t without knowing the indefinite integral of sin(x).