f'(x) ＝2x-2a
∵f(x)在x＝1处有极值2
∴f'(1)=0,且f(1)=2
∴2-2a=0,1-2a+b=2
∴a=1,b=3
f(x)=x^2-2x+3=(x-1)^2+2
∵x∈[0,3]
∴x=1,f(x)min=2
x=3,f(x)max=6
2
y=x^2-2x+3与y=x+3
==>x^2-3x=0,
==>x1=0,x2=3
∴曲线y＝x^2－2ax＋b,y＝x＋3所围成的图形的面积
S=ʃ(0,3)[x+3-x^2+2x-3)dx
=ʃ(0,3)[-x^2+3x)dx
=(-1/3*x^3+3/2*x^2)|(0,3)
=9/2