∫ − 1 0 ( 2 x − e x ) d x = 2 x 2 2 − e x | − 1 0 = − 1 − ( 1 − 1 e ) = 1 e − 2 {\displaystyle \int _{-1}^{0}(2x-e^{x})dx={\frac {2x^{2}}{2}}-e^{x}{\bigg |}_{-1}^{0}=-1-(1-{\frac {1}{e}})={\frac {1}{e}}-2}
