Question

integtal 0 до -2 (x^2e^(-x/2))

Answer 1

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