微積分問題 - 考試
By John
at 2014-02-24T11:35
at 2014-02-24T11:35
Table of Contents
出處 周易工程數學
∫ (x+1)^2 dx
我的想法 ∫ (x+1)^2 d(x+1) =1/3 (x+1)^3+ c
另解
(x+1)^2 = x^2+2x+1
∫ x^2+2x+1 dx = 1/3x^3+x^2+x+c1.....(1)
1/3(x+1)^3+ c = 1/3(x^3+3x^2+3x+1)+c=1/3x^3+x^2+x+1/3+c...(2)
照理說(1)應該等於(2)
但差了1/3
題目有給初值所以應該不是c1=1/3+c
請版上大大解惑
--
∫ (x+1)^2 dx
我的想法 ∫ (x+1)^2 d(x+1) =1/3 (x+1)^3+ c
另解
(x+1)^2 = x^2+2x+1
∫ x^2+2x+1 dx = 1/3x^3+x^2+x+c1.....(1)
1/3(x+1)^3+ c = 1/3(x^3+3x^2+3x+1)+c=1/3x^3+x^2+x+1/3+c...(2)
照理說(1)應該等於(2)
但差了1/3
題目有給初值所以應該不是c1=1/3+c
請版上大大解惑
--
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