更新1:
Answer is 1.
Definite Integral
回答 (2)
2012-05-24 3:17 am
✔ 最佳答案
∫_[-π/2~0] x sinx/(1+e^x) dx +∫_[0~π/2] x sinx/(1+e^x)dx=∫_[π/2~0] -xsin(-x)/[1+e^(-x)] (-dx)+ ∫_[0~π/2] xsinx/(1+e^x) dx
=∫_[0~π/2] xsinx e^x/(e^x+1) dx +∫_[0~π/2] xsinx/(1+e^x)dx
=∫_[0~π/2] xsinx dx
= -xcosx+ sinx for x=0~π/2
= 1
2012-05-24 4:10 pm
媽弗勒時﹗媽弗勒時﹗
2012-05-24 08:10:17 補充：
媽弗勒時﹗媽弗勒時﹗
2012-05-24 08:10:17 補充：
媽弗勒時﹗媽弗勒時﹗
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