Evaluate the integral
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Calculus
2015-03-31 7:32 pm
回答 (2)
2015-03-31 10:58 pm
2015-03-31 8:53 pm
∫(√y－y)/y^2 dy [y＝1 to 4]＝∫(y^(-3/2)－y^(-1)] dy＝∫(y^(-3/2)－y^(-1)] dy＝[-2y^(-1/2)－ln y]＝[-2/√4－ln 4] - [-2/√1－ln 1]＝-1－2 ln 2＋2＝1－2 ln 2
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https://hk.answers.yahoo.com/question/index?qid=20150331000051KK00035