求得3个交点为(1,1)(0.5,2)(2,2)。按理说用Y型来求,这个没问题,可是用X型,我怎么也求不出来,附上我的计算,麻烦指出错误。答案是19/6
0.5<=x<=1,1/x<=y<=2 另外1<=x<=2,x<=y<=2
原式=∫[0.5,1] {∫[1/x,2](2x+y)dy}dx+∫[1,2]{∫[x,2](2x+y)dy}dx
=∫[0.5,1](4x-1/2x^2)dx+∫[1,2](4x-2.5x^2+2)dx
=(2x^2+1/2x)|[1,0.5]+(2x^2-5x^3/6+2x)|[2,1]
=9/6