求抛物线y^2=2x与直线y=4-x围成平面图形的面积?

1.先求抛物线与直线的交点
y^2=2x y=4-x
(4-x)^2=2x
x^2-10x+16=0
x1=2 y1=4-2=2 点(2,2)
x2=8 y2=4-8=-4 点(8,-4)
2.再求积分 y积分范围从-4到2(上2,下-4,下同)
y^2=2x x=y^2/2
y=4-x x=4-y
∫(-4,2)(4-y-y^2/2)dy
=(4y-1/2y^2-y^3/6)|(-4,2)
=(8-2-4/3)-(-16-8+32/3)
=30-12
=18