Question

计算二次积分 ∫（0到1）dy∫（y到1）sinx/xdx

Answer 1

解：由题意分析知，此二次积分的积分区域是以(0,0)和(1,0)和(1,1)三点为顶点的直角三角形区域
故∫(0,1)dy∫(y,1)(sinx/x)dx=∫(0,1)(sinx/x)dx∫(0,x)dy (变换积分顺序)
=∫(0,1)(sinx/x)[y│(0,x)]dx
=∫(0,1)(sinx/x)(x-0)dx
=∫(0,1)sinxdx
=-cosx│(0,1)
=-cos1+cos0
=1-cos1

Answer 2

交换积分次序得
∫[0,1]dy∫[y,1]sinx/xdx
=∫[0,1]dx∫[x,1]sinx/xdy
=∫[0,1]dxysinx/x[x,1]
=∫[0,1](sinx/x-sinx)dx
∫[0,1]sinx/xdx是一个变异积分啊