$\displaystyle y = \sqrt{2 - x^{2}}$ evaluated at upper bound 1 and lower bound 0
$\displaystyle y'(\sqrt{2 - x^{2}}) = -2x$
$\displaystyle L = \int_{0}^{1} \sqrt{1 + (-2x)^{2}}$
$\displaystyle L = \int_{0}^{1} \sqrt{1 + 4x^{2}}$
$\displaystyle L = \int_{0}^{1} (1 + 4x^{2})^{1/2}$
$\displaystyle L = \int_{0}^{1} (u)^{1/2}$
$\displaystyle u = 1 + 4x^{2}$
$\displaystyle du = 8x$
$\displaystyle \dfrac{1}{8}du = dx$
$\displaystyle L = \int_{0}^{1} (u)^{1/2}$
$\displaystyle L = \dfrac{1}{8} \int_{0}^{1} ((u)^{1/2}$
$\displaystyle (\dfrac{2}{3})(\dfrac{1}{8})(u)^{3/2}$ evaluated at upper bound 1 and lower bound 0
$\displaystyle (\dfrac{1}{12})(u)^{3/2}$ evaluated at upper bound 1 and lower bound 0
$\displaystyle (\dfrac{1}{12})(1 + 4x^{2})^{3/2}$ evaluated at upper bound 1 and lower bound 0
$\displaystyle [(\dfrac{1}{12})(1 + 4(1)^{2})^{3/2}] - (\dfrac{1}{12})(1 + 4(0)^{2})^{3/2}$ - Is this right?