应该是(0,-3)
y'=-2x+4
x=0,k=y'=4
x=3,k=y'=-2
所以切线是y+3=4(x-0),y-0=-2(x-3)
即y=4x-3和y=-2x+6
交点是(3/2,3)
所以面积=∫(0到3/2)[4x-3-(-x²+4x-3)]dx+∫(3/2到3)[-2x+6-(-x²+4x-3)]dx
=(x³/3)(0到3/2)+(x³/3-3x²+9x)(3/2到3)
=(9/8)+(9-63/8)
=9/4