Ответ:

[tex] \displaystyle \int_{1}^{2} x \times {e}^{ - x} dx[/tex]

[tex]u = x \\dv = {e}^{ - x} [/tex]

[tex]du = x' \\ v = \int {e}^{ - x} [/tex]

[tex]du = 1 \\ v = - {e}^{ - x} [/tex]

Формула интегрирования по частям:

[tex] \displaystyle \int u \ dv = u \ v - \int v \ du[/tex]

[tex] \displaystyle \int_{1}^{2}x \times {e}^{ - x} dx = x \times ( - {e}^{ - x} ) - \int - {e}^{ - x} \times 1dx = ( - x \times {e}^{ - x} - {e}^{ - x} )| _{1}^{2} = - 2 \times {e}^{ - 2} - {e}^{ - 2} - ( - 1 \times {e}^{ -1} - {e}^{ - 1} ) = - 3 {e}^{ - 2} - ( - 2 {e}^{ - 1} ) = - \frac{3}{ {e}^{2} } + \frac{2}{e} = \frac{ - 3 + 2e}{ {e}^{2} } [/tex]