Find the area of the region R bounded by y= x^{6}/3 4, the xaxis, x= 1, and x= 1.
Calculus

Tutor: None Selected  Time limit: 1 Day 
Find the area of the region bounded by , the axis, , and . How would I solve this?
given y=x^{6}/34 between x=1 and x=1 using integral calculus method
integrating y with respect to X gives area (A),when we substitute boundary conditions
the integral will be A= [x^{7}/214x+c] where c=arbitrary constant, A= area in square units
but taking boundary conditions x= 1 and 1 and substituting into equation,
A= [(1/214)(1/21+4)]
area = 8 square units ( N/B negative sign is assumed for area)
used integral calculus method to solve it. arbitrary constant will cancel for an homogenous integrals due to boundary given.
How do you account the vertical yaxis values where the minimum is 4 to xaxis?
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 XAxis boundary accounted for?
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 yaxis it is 5.18 and w/ respect of xaxis it is 7.905. Why am I getting two different values altogether?
good argument and reasoning
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
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
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
Secure Information
Content will be erased after question is completed.