先求交点：t-t^3=1-t^4 t1=1 t2=-1 dA=(1/2)*∮(xdy-ydx)=0.5∮[（t-t^3)（-4t^3)-(1-t^4)((1-3t^2)]dt =0.5∮[（t-t^3)（-4t^3)-(1-t^4)((1-3t^2)]dt =0.5∮[（-t^4+2t^2-1)]dt =0.5*2∫（0,1）(1-2t^2+t^4)dt=(t-2t^3/3+t^5/5)|(0,1)=8/15