y^2=2x+1,y=x-1 求这两条曲线围成的平面图形的面积画出图像 并写出具体步骤 谢谢!
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=∫[-1,3](y+1-y^2/2+1/2)dy
=∫[-1,3](y-y^2/2+3/2)dy
=[y^2/2-y^3/6+3/2 x]\[-1,3]
=9/2-1/2-27/6-1/6+6
=10-14/3
=16/3