(1);原式=Ln(x^2+1)对他求定积分得Ln(2^2+1)-Ln(0^2+1)=Ln5 (2):原式=x(lnx-1)所以是定积分e(lne-1)-e(ln1-1)=e (3):原式=[sin(2x)]/2,求定积分得[sin(2π/4)]/2-[sin(0)]/2=0.5
(1);原式=Ln(x^2+1)对他求定积分得Ln(2^2+1)-Ln(0^2+1)=Ln5 (2):原式=x(lnx-1)所以是定积分e(lne-1)-e(ln1-1)=e (3):原式=[sin(2x)]/2,求定积分得[sin(2π/4)]/2-[sin(0)]/2=0.5