How do I compute integral of t to the power of 2 between 3 and 1?
回答 (2)
2013-05-01 10:36 am
∫ t² dt = [ t³/3 ]__________limits 1 to 3
9 - 1/3 = 26/3 = 8 ⅔
9 - 1/3 = 26/3 = 8 ⅔
2013-05-01 10:36 am
∫ (t = 1 to 3) t^2 dt
= [1/3 * t^3] (t = 1 to 3)
= 1/3 * 3^3 - 1/3 * 1^3
= 26/3
= [1/3 * t^3] (t = 1 to 3)
= 1/3 * 3^3 - 1/3 * 1^3
= 26/3
收錄日期: 2021-05-01 14:46:54
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https://hk.answers.yahoo.com/question/index?qid=20130501023119AA63i97