二曲线方程联立,解出二线交点
A(a/2,2a),B(2a,a/2),
积分区域:a/2答案是: (15/8-ln4)a^2??二曲线方程联立,解出二线交点A(a/2,2a),B(2a,a/2),积分区域:a/2<=x<=2a, a^2/x<=y<=5a/2-x,I=∫ [a/2,2a] dx∫ [a^2/x,-x+5a/2]dy=∫ [a/2,2a] ( 5a/2-x-a^2/x)dx= [a/2,2a]( 5ax/2-x^2/2-a^2ln|x|)=(5a^2-2a^2-a^ln(2a)-(5a^2/4-a^2/8-a^2ln(a/2)=3a^2-a^2ln(2a)-9a^2/8+a^2ln(a/2)=15a^2/8-a^2(ln2+lna+lna-ln2)=15a^2/8-2a^2ln2.