Question

这个高数解题过程，

它俩咋就等于零了呢

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

Answer 1

（3）原式=∫(1,2)dy∫(y,y^2)sin(πx/2y)dx =∫(1,2)dy*[-(2y/π)*cos(πx/2y)|(y,y^2)] =∫(1,2) (-2y/π)*cos(πy/2)dy =∫(1,2) (-4y/π^2)*d[sin(πy/2)] =(-4y/π^2)*sin(πy/2)|(1,2)+∫(1,2) (4/π^2)*sin(πy/2)dy =4/π^2-(8/π^3)*cos(πy/2)|(1,2) =4/π^2+8/π^3