Question

高数积分题目 求解

Answer 1

解：所求面积=│∫<-2,-1>[(2x+2)-(2-x²)]dx│+∫<-√2,-1>(2-x²)dx+∫<-1,0>[(2-x²)-(2x+2)]dx
(说明：∵第一个积分所表示的面积图形负y轴上，它是负值∴应取绝对值)
=│∫<-2,-1>(x²+2x)dx│+∫<-√2,-1>(2-x²)dx-∫<-1,0>(x²+2x)dx
=│(x³/3+x²)│<-2,-1>│+(2x-x³/3)│<-√2,-1>-(x³/3+x²)│<-1,0>
=│(-1/3+1)-(-8/3+4)│+[(-2+1/3)-(-2√2+2√2/3)]-[(0+0)-(-1/3+1)]
=│-2/3│+(4√2/3-5/3)-(-2/3)
=(4√2-1)/3

Answer 2

先求两者交点(0,2)，(-2，-2)
∫[-2,2][-√(2-y)-(y-2)/2]dy
=[2/3*(2-y)^(3/2)-y^2/4+4y][-2,2]
=-1+8-16/3+1+8
=8/3

Answer 3

2 - x^2 = 2 x + 2
得两个交点为x=-2和x=0
于是面积为∫|2 - x^2 - 2 x + 2|dx (从-2到0)
=4/3