What is the arclength of $f(x)=(x^2-2x)/(2-x)$ on $x in [-2,-1]$ ?

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Answer 1 · 2017-12-30T12:45:01

Arc length of $f(x)$ is $sqrt2$ unit.

$f(x) = (x^2-2x)/(2-x) ; x in [-2,-1]$

$f'(x) = ((2x-2)(2-x)-(x^2-2x)(-1))/(2-x)^2$ or

$f'(x) = ((2x-2)(2-x)+(x^2-2x))/(2-x)^2$ or

$f'(x) = ((2x-2)(2-x)+x(x-2))/(2-x)^2$ or

$f'(x) = ((2-x)(2x-2-x))/(2-x)^2$ or

$f'(x) = ((2-x)(x-2))/(2-x)^2=(x-2)/(2-x)=-1$

$:. [f'(x)]^2=1$ . Total length of the arc from x = a to x = b is

$int_a^b sqrt(1+[f'(x)]^2)dx$

$L=int_(-2)^(-1) sqrt(1+1)dx$ or

$L=int_(-2)^(-1) sqrt(2) dx$ or

$L=sqrt2 [x]_-2^-1or L= sqrt2[-1-(-2)]$ or

$L= sqrt2[-1+2] =sqrt2$

Arc length of $f(x)$ is $sqrt2$ unit. [Ans]

What is the arclength of f(x)=(x^2-2x)/(2-x)f(x)=(x^2-2x)/(2-x) on x in [-2,-1]x in [-2,-1]?