$\displaystyle y = \sqrt{2 - x^{2}}$ 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?