find the exact area of the surface by rotating curve about the x-axis. y=x^3/3+1/(4x), 1/2 < or equal to x < or equal to 1 please be thorough

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You should remember the area formula for a surface generated by rotating the curve f(x) around x axis such that:

`A = 2pi*int_a^b f(x)*sqrt(1 + (f'(x))^2) dx`

Hence, you need to substitute `1/2`  for a and 1 for b such that:

`A = 2pi*int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx`

You should find the derivative of the function such that:

`f'(x) = 3x^2/3 - 4/(16x^2) => f'(x) = x^2 - 1/(4x^2)`

Squaring the derivative yields:

`(f'(x))^2 = x^4 - 2x^2*(1/4x^2) + 1/(16x^4)`

`(f'(x))^2 = x^4 - 1/2 + 1/(16x^4)`

Hence, evaluating `1 + (f'(x))^2`  yields:

`1 + (f'(x))^2 = 1 + x^4 - 1/2 + 1/(16x^4)`

`1 + (f'(x))^2 = x^4+ 1/2 + 1/(16x^4) = (x^2+ 1/(4x^2))^2`

Hence, taking the square root of `1 + (f'(x))^2`  yields:

`sqrt((x^2 + 1/(4x^2))^2) = (x^2 + 1/(4x^2))`

You need to multiply `sqrt((x^2 + 1/(4x^2))^2)`  by f(x) to find the integrand such that:

`(x^2 + 1/(4x^2))*(x^3/3 + 1/(4x)) = x^5/3 + x/4 + x/12 + 1/(16x^3)`

`(x^2 + 1/(4x^2))*(x^3/3 + 1/(4x)) = x^5/3 + x/3 + 1/(16x^3)`

Hence, using the property of linearity of integral yields:

`int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = int_(1/2)^1 (x^5/3) dx+ int_(1/2)^1 x/3 dx+ int_(1/2)^1 1/(16x^3) dx`

`int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = ((1/3)(x^6/6) + (1/3)(x^2/2)- (1/32)(1/x^2))|_(1/2)^1 `

`int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = (1/18)(1 - 1/64) + 1/6(1 - 1/4) - (1/32)(1 - 4)`

`int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = 63/(18*64) + 3/(6*4) + 3/32`

`int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = 7/128 + 1/8 + 3/32`

`int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = (7 + 16 + 12)/128`

`int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = 35/128`

`A = 2pi*int_(1/2)^1 f(x)*sqrt(1 + (f'(x))^2) dx = 35pi/64`

Hence, evaluating the area of surface of revolution of curve f(x), about x axis, yields `A= 35pi/64` .

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