如题所述
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
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
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