f(x)=x^3 + ax^2 + bxf'(x)=3x^2+2ax+b3+2a+b=0 且 1+a+b=-22a+b=-3 且 a+b=-3解得：a=0 b=-3S=∫[-√3,0](x^3-3x)dx+∫[0,√3](3x-x^3)dx =[x^4/4-3x^2/2]|[-√3,0]+[3x^2/2-x^4/4]|[0,√3] =9/4+9/4 =9/2 =4.5
