计算由抛物线y^2=2x和直线y=x-4 所围成的图形的面积.

y^2=2x,---->x=y^2/2y=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...