定积分的计算

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2 | 2012-8-28 03:19:28 | 显示全部楼层 |阅读模式

题目不太清楚，如果(-x/5)是指数，则: ∫ [0->+∞][e^(-x/5)]dx = (-5) ∫ [0->+∞][e^(-x/5)]d(-x/5)= (-5)e^(-x/5) [0->+∞]= (-5)(0 - 1) = 5如果(-x)是指数，则: ∫ [0->+∞][e^(-x)/5]dx = (-1/5) ∫ [0->+∞]e^(-x)d(-x)= (-1/5)e^(-x) [0->+∞]= (-1/5)(0 - 1) = 1/5...

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