高等数学:应用定积分求由下列曲线围成的平面的面积
y^2=2x+1,y=x-1
x^2-2x+1=2x+1
x^2=4x
x1=0？x2=4？y1=0？y2=3
s=∫(y+1-y^2/2+1/2)dy
=∫(y-y^2/2+3/2)dy
=[y^2/2-y^3/6+3/2 x]\
=9/2-1/2-27/6-1/6+6
=10-14/3
=16/3
x^2-2x+1=2x+1
x^2=4x
x1=0？x2=4？y1=0？y2=3
s=∫(y+1-y^2/2+1/2)dy
=∫(y-y^2/2+3/2)dy
=[y^2/2-y^3/6+3/2 x]\
=9/2-1/2-27/6-1/6+6
=10-14/3
=16/3