设f(x)=
{xe^-x2，x≥0，
{1/(1+cosx)，-1≤x＜0，
求∫₁⁴f(x-2)dx.

设f(x)=
{xe^-x2，x≥0，
{1/(1+cosx)，-1≤x＜0，
求∫₁⁴f(x-2)dx.

令x-2=t，则 ∫₁⁴f(x-2)dx=∫_-1²f(t)dt =∫_-1⁰[1/(1+cost)]dt+∫₀²te^-t2dt =1/2∫_-1⁰[1/cos²(t/2)]dt-1/2∫₀²e^-t2d(-t²) =tan(t/2)|_-1⁰-(1/2)e^-t2|₀²=tan(1/2)-(1/2)e^-4+1/2．