一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
设函数 $f(t)$ 连续,令 $F(x, y)$ $=$ $\int_{0}^ {x-y}(x-y-t) f(t)\mathrm {~d} t$, 则( )
A. $\frac { \partial F } { \partial x }$ $=$ $\frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $\frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$
B. $\frac { \partial F } { \partial x }$ $=$ $\frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $- \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$
C. $\frac { \partial F } { \partial x }$ $=$ $- \frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $\frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$
D. $\frac { \partial F } { \partial x }$ $=$ $- \frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $- \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$
难度评级:
二、解析 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
首先,对题目所给的式子进行变形:
$$
\begin{aligned}
F ( x , y ) \\ \\
& = \int _{ 0 } ^ { x – y } ( x – y – t ) f ( t ) \mathrm{ ~d } t \\ \\
& = ( x – y ) \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – \int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t \\ \\
& = x \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – y \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – \int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t
\end{aligned}
$$
于是,对 “$\textcolor{springgreen}{x}$” 求偏导的时候,”$\textcolor{yellow}{y}$” 与 “$\textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}$” 都要当作常数处理:
$$
\begin{aligned}
\frac { \partial F } { \partial x } \\ \\
& = \left[ \textcolor{springgreen}{x} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{y \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{\int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t} \right]^{\prime}_{\textcolor{springgreen}{x}} \\ \\
& = \left[ \textcolor{springgreen}{x} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \right] ^{\prime}_{x} \\ \\
& = \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}
\end{aligned}
$$
类似的,对 “$\textcolor{springgreen}{y}$” 求偏导的时候,”$\textcolor{yellow}{x}$” 与 “$\textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}$” 都要当作常数处理:
$$
\begin{aligned}
\frac { \partial F } { \partial y } \\ \\
& = \left[ \textcolor{yellow}{x \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{springgreen}{y} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{\int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t} \right] ^{\prime}_{\textcolor{springgreen}{y}} \\ \\
& = \left[ – \textcolor{springgreen}{y} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \right] ^{\prime} _{\textcolor{springgreen}{y}} \\ \\
& = \textcolor{yellow}{- \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}
\end{aligned}
$$
于是可知:
$$
\textcolor{magenta}{
\boldsymbol{
\frac { \partial F } { \partial x } = – \frac { \partial F } { \partial y }
}
}
$$
于是可以排除 [A] 和 [B] 两个选项。
接着,由 $\frac { \partial F } { \partial x }$ $=$ $\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t$ 可得:
$$
\begin{aligned}
\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } } \\ \\
& = \frac{\partial}{\partial x} \left( \frac { \partial F } { \partial x } \right) \\ \\
& = \left( \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t \right) ^{\prime} _{x} \\ \\
& = \textcolor{orangered}{x ^{\prime}} + f(x – y) \\ \\
& = f ( x – y )
\end{aligned}
$$
类似的,由 $\frac { \partial F } { \partial y }$ $=$ $- \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t$ 可得:
$$
\begin{aligned}
\frac { \partial ^ { 2 } F } { \partial y ^ { 2 } } \\ \\
& = \frac{\partial}{\partial y} \left( \frac { \partial F } { \partial y } \right) \\ \\
& = \left( – \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t \right) ^{\prime} _{y} \\ \\
& = \textcolor{orangered}{-(-y) ^{\prime}} + f(x – y) \\ \\
& = f ( x – y )
\end{aligned}
$$
于是可知:
$$
\textcolor{magenta}{
\boldsymbol{
\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }
}
}
$$
综上可知,本 题 应 选 C
高等数学![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
线性代数![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
以独特的视角解析线性代数,让繁复的知识变得直观明了。
特别专题![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。