Find the Arc Length...Calculus

Find arc length of y = sqrt{1-x} on the interval [0,4].

My answer is 6. Is this right?

y = ( 1 - x )^( 1 / 2 )
dy / dx = (- 1 / 2 )(1 - x)^( - 1 / 2 )

int [ sqrt [ 1 + [ ( - 1 / 2 )(1 - x)^( - 1 / 2 ) ]^2 ] dx ]
= int [ sqrt [ 1 + (1 / 4 )(1 - x)^( - 1 ) ] dx ]
= int [ sqrt [ [( 1 - x) + 1 ] / 4(1 - x ) ] dx ]
= int [ sqrt [ ( 2 - x) / ( 4 - 4x ) ] dx ]
= (1 / 2 )int [ sqrt ( 2 - x ) / sqrt ( 1 - x ) dx ]
Let 1 - x = tan^2 u
- dx = 2 tan u sec^2 u du
dx = - 2 tan u sec^2 u du

= ( 1 / 2 ) * int [ sqrt ( 1 + tan^2 u ) * - 2 tan u sec^2 u du / (sqrt ( tan^2 u ) ]
= - 2 *( 1 / 2 ) * int [ sec^3 u tan u du / tan u ]
= - int [ sec^3 u du ]


int [sec^3 u du ]= [ sec u tan u - int [ sec u tan^2 u dx ] ]
int [ sec^3 u du ] = [ sec u tan u - int [ sec u ( sec^2 u - 1 ) dx ] ]
int [ sec^3 u du ] = sec u tan u - int [ sec^3 u du ] + int [ sec u du ]
2 * int [ sec^3 u du ] = sec u tan u + ln abs ( sec u + tan u )
int [ sec^3 u du ] = ( 1 / 2 ) [ sec u tan u + ln abs ( sec u + tan u ) ]

= ( - 1 / 2 ) [ sec u tan u + ln abs ( sec u + tan u ) ]

Then replace the trig functions with functions of x.

y = sqrt(1 - x) is not real-valued on [0,4], since (1 - x) < 0 for x > 1, so there
is no solution over the field of real numbers.

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