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?

simple, initially your equation was function of x right? that is y(x) but you decided to express x in terms of y giving you x as function of y i.e x(y) right and you got different answers well,

if you express in terms of y, then boundary conditions will never be x=-4 or 0, you use values of y on given point.

example, points (-4,y1) and (0,y2) so your limits will be y1 and y2

thank you

Apr 26th, 2015

N/B given boundary values as real values of x and function in terms of x then integrate with respect to x

and if you decide otherwise by expressing function in terms of y, then use boundary values as real values of y at same given point

Apr 26th, 2015

N/B given boundary values as real values of x and function in terms of x then integrate with respect to x

and if you decide otherwise by expressing function in terms of y, then use boundary values as real values of y at same given point

Apr 26th, 2015

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