The volume of the solid obtained by rotating the region bounded by the curves `y=sqrt(x-1), y=0, x= 5` , about x axis, can be evaluated using the washer method, such that:

`V = int_a^b pi*(f^2(x) - g^2(x))dx`

Since the problem provides you the endpoint x = 5, you need to...

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The volume of the solid obtained by rotating the region bounded by the curves `y=sqrt(x-1), y=0, x= 5` , about x axis, can be evaluated using the washer method, such that:

`V = int_a^b pi*(f^2(x) - g^2(x))dx`

Since the problem provides you the endpoint x = 5, you need to find the other endpoint of interval, hence, you need to solve for x the following equation, such that:

`sqrt (x-1) = 0 => x - 1 = 0 => x = 1`

`V = int_1^5 pi*(sqrt(x-1) - 0)^2 dx`

`V = pi*int_1^5 (x - 1)dx`

`V = pi*int_1^5 (x)dx - pi*int_1^5 dx`

`V = (pi*x^2/2 - pi*x)|_1^5`

`V = (pi*5^2/2 - pi*5 - pi*1^2/2 + pi*1)`

`V = (25pi)/2 - pi/2 - 4pi`

`V = 12pi - 4pi`

`V = 8pi`

**Hence, evaluating the volume of the solid obtained by rotating the region bounded by the curves `y=sqrt(x-1), y=0, x= 5` , about x axis , using the washer method, yields `V = 8pi.` **