第六题，感谢
答案:1 悬赏:40 手机版
解决时间 2021-02-02 16:55
- 提问者网友：孤凫
- 2021-02-02 13:33
第六题，感谢
最佳答案
- 五星知识达人网友：酒者煙囻
- 2021-02-02 14:40
令x=-u，则dx=-du，u：1--->-1
∫[-1-->1] x²/(1+e^x)dx
=-∫[1-->-1] u²/(1+e^(-u))du
=∫[-1-->1] u²e^u/(e^u+1)du
积分变量可以随便换字母
=∫[-1-->1] x²e^x/(e^x+1)dx
则得到：∫[-1-->1] x²/(1+e^x)dx=∫[-1-->1] x²e^x/(e^x+1)dx
因此
∫[-1-->1] x²/(1+e^x)dx
=1/2[ ∫[-1-->1] x²/(1+e^x)dx + ∫[-1-->1] x²e^x/(e^x+1)dx ]
=1/2∫[-1-->1] [ x²/(1+e^x)+x²e^x/(e^x+1) ]dx
=1/2∫[-1-->1] x² dx
=(1/6)x³ |[-1-->1]
=(1/6)-(-1/6)
=1/3
∫[-1-->1] x²/(1+e^x)dx
=-∫[1-->-1] u²/(1+e^(-u))du
=∫[-1-->1] u²e^u/(e^u+1)du
积分变量可以随便换字母
=∫[-1-->1] x²e^x/(e^x+1)dx
则得到：∫[-1-->1] x²/(1+e^x)dx=∫[-1-->1] x²e^x/(e^x+1)dx
因此
∫[-1-->1] x²/(1+e^x)dx
=1/2[ ∫[-1-->1] x²/(1+e^x)dx + ∫[-1-->1] x²e^x/(e^x+1)dx ]
=1/2∫[-1-->1] [ x²/(1+e^x)+x²e^x/(e^x+1) ]dx
=1/2∫[-1-->1] x² dx
=(1/6)x³ |[-1-->1]
=(1/6)-(-1/6)
=1/3
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