Using Property (1) of the Definite Integral

\[ \begin{eqnarray*} \int_{0}^{1}(x^{2}+e^{x})\,dx &=& \int_{0}^{1}x^{2}dx+\int_{0}^{1}e^{x}\,dx= \left[ \dfrac{x^{3}}{3}\right] _{0}^{1}+ \Big[ e^{x}\Big] _{0}^{1} \\[4pt] &=&\left[ \dfrac{1^{3}}{3}-0\right] +\Big[ e^{1}-e^{0}\Big] =\dfrac{1}{3}+e-1=e-\dfrac{2}{3} \end{eqnarray*} \]