求抛物线y平分=2x与直线y=4-x围成的平面图形的面积。

Question

如题所述

Answer 1

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

Answer 2

解： y^2=2x,---->x=y^2/2
y=x-4,---->x=y+4.
y^2=2x与y=x-4的交点是(2,-2)(8,4)
所围成的图形的面积=∫(4,-2),[(y+4)-y^2/2]dy=[y^2/2+4y-y^3/6],(4,-2)
=(4^2/2+4*4-4^3/6)-[(-2)^2/2-4*2-(-2)^3/6]
=8+16-32/3-2+8-4/3
=18