f(x)=x^3 + ax^2 + bx
f'(x)=3x^2+2ax+b
3+2a+b=0 且 1+a+b=-2
2a+b=-3 且 a+b=-3
解得：a=0 b=-3
S=∫[-√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
