\[\sqrt{\frac{m}{2\alpha}}\left(\frac{y^{1/2}}{1/2}\right)_{\frac{E}{\alpha}+x_{0}}^{\frac{E}{\alpha}+x}=\sqrt{\frac{2m}{\alpha}}\left[\left(\frac{E}{\alpha}+x\right)^{1/2}-\left(\frac{E}{\alpha}+x_{0}\right)^{1/2}\right].\] Então, temos