for such you will only take boundary conditions as x=-4 and x=0, in other wards you consider from 0 down to negative integers. otherwise your equation would not make sense if you substitute directly negative values of x with positive ones.
basically you're saying to take the integration (1/3)*x^6 - 4 w/ respect of x in interval from 0 to 1 multiplied by 2. How is the X-Axis boundary accounted for?
Apr 25th, 2015
Because for whatever the reason, I did exactly that, and the integration of x = (3*y + 12)^(1/6) with respect of y w/ interval from -4 to 0. I have two different answers. w/ respect of y-axis it is 5.18 and w/ respect of x-axis it is -7.905. Why am I getting two different values altogether?