Question

integtal 0 до -2 (x^2e^(-x/2))

Answer 1

$displaystyleintlimits^0_{-2}x^2e^{-frac{x}{2}}dx=-2x^2e^{-frac{x}{2}}|^0_{-2}+4intlimits^0_{-2} xe^{-frac{x}{2}}dx=\=-2x^2e^{-frac{x}{2}}|^0_{-2}+4(-2xe^{-frac{x}{2}}|^0_{-2}+2intlimits^0_{-2} e^{-frac{x}{2}}dx)=\=-2x^2e^{-frac{x}{2}}|^0_{-2}-8xe^{-frac{x}{2}}|^0_{-2}-16e^{-frac{x}{2}}|^0_{-2}=\=-16+8e-16e+16e=8e-16\\u_1=x^2;du_1=2xdx\dv_1=e^{-frac{x}{2}}dx;v_1=-2e^{-frac{x}{2}}\u_2=x;du_2=dx\dv_2=e^{-frac{x}{2}}dx;v_2=-2e^{-frac{x}{2}}$