Answer
$L = \int^{0}_{2} \sqrt{1+(2y-2)^{2}} dy$
Work Step by Step
$x = y^{2} - 2y$
Then $dx/ dy = 2y - 2$
and $1+(dx/dy)^2 = 1+(2y-2)^{2}$
Therefore
$L = \int^{0}_{2} \sqrt{1+(2y-2)^{2}} dy$
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