变上限积分定义的第二个推论(B007)

问题

若函数 $\textcolor{Orange}{f(x)}$ 在区间 $[a, b]$ 上连续,函数 $\textcolor{Orange}{\phi(x)}$ 和 函数 $\textcolor{Orange}{\mu(x)}$ 在区间 $[a, b]$ 上可导,且变限积分 $\textcolor{Orange}{F(x)}$ $\textcolor{Orange}{=}$ $\textcolor{Orange}{{\int}_{\mu(x)}^{\phi(x)}}$ $\textcolor{Orange}{f(t)}$ $\textcolor{Orange}{\mathrm{d} t}$, 则 $\textcolor{Orange}{F^{\prime}(x)}$ $=$ $?$

选项

[A].   $F^{\prime}$ $=$ $f[\phi(x)]$ $\phi(x)$ $-$ $\mu(x)$ $\mu(x)$

[B].   $F^{\prime}$ $=$ $f[\phi(x)]$ $\phi^{\prime}(x)$ $+$ $\mu(x)$ $\mu^{\prime}(x)$

[C].   $F^{\prime}$ $=$ $f[\phi(x)]$ $\phi^{\prime}(x)$ $-$ $\mu(x)$ $\mu^{\prime}(x)$

[D].   $F^{\prime}$ $=$ $f[\phi(x)]$ $\phi(x)$ $+$ $\mu(x)$ $\mu(x)$


上一题 - 荒原之梦   答 案   下一题 - 荒原之梦

$$F^{\textcolor{Yellow}{\prime}} =$$ $$\Bigg[ \int_{\textcolor{Orange}{\mu(x)}}^{\textcolor{Orange}{\phi(x)}} f(t) \mathrm{d} t \Bigg]^{\textcolor{Yellow}{\prime}} =$$ $$f[\textcolor{Orange}{\phi(x)}] \textcolor{Green}{\cdot} \textcolor{Orange}{\phi} ^{\textcolor{Yellow}{\prime}} \textcolor{Orange}{(x)}$$ $$\textcolor{Green}{-}$$ $$f[\textcolor{Orange}{\mu(x)}] \textcolor{Green}{\cdot} \textcolor{Orange}{\mu} ^{\textcolor{Yellow}{\prime}} \textcolor{Orange}{(x)}.$$


荒原之梦网全部内容均为原创,提供了涵盖考研数学基础知识、考研数学真题、考研数学练习题和计算机科学等方面,大量精心研发的学习资源。

豫 ICP 备 17023611 号-1 | 公网安备 - 荒原之梦 豫公网安备 41142502000132 号 | SiteMap
Copyright © 2017-2024 ZhaoKaifeng.com 版权所有 All Rights Reserved.

Copyright © 2024   zhaokaifeng.com   All Rights Reserved.
豫ICP备17023611号-1
 豫公网安备41142502000132号

荒原之梦 自豪地采用WordPress