求由曲线y^2=2x,与直线y=x-4所围成的图形D的面积

解法一：图形D的面积=∫(-2,4)(y+4-y²/2)dy
=(y²/2+4y-y³/6)│(-2,4)
=4²/2+4*4-4³/6-(-2)²/2-4*(-2)+(-2)³/6
=18
解法二：图形D的面积=∫(0,2)2√(2x)dx+∫(2,8)[√(2x)-x+4]dx
=[(4√2/3)x^(3/2)]│(0,2)+[(2√2/3)x^(3/2)-x²/2+4x]│(2,8)
=16/3++64/3-32+32-8/3+2-8
=18