\documentclass[../main.tex]{subfiles}
\begin{document}
\chapter{2023-2024学年微积分（一）（下）期中考试}

\section{基本计算题(每小题 6 分，共 60 分)}
\begin{enumerate}
    \item 已知$\boldsymbol{a}\times (\boldsymbol{b}\times \boldsymbol{a}) = \boldsymbol
              {b}-2\boldsymbol{a}$，且$|\boldsymbol{a}|=1$，$|\boldsymbol{b}|=4$，求$|\boldsymbol
              {b}+\boldsymbol{a}|$.
          \vspace{10em}

    \item 已知单位向量$\overrightarrow{OA}$与$x$轴正向的夹角为$\frac{\pi}{3}$，与$y$轴正向的夹角为$\frac{\pi}{4}$，且在$z$轴上的坐标是负的.
          $\overrightarrow{OB}=\left(1,-\sqrt{2},-1\right)$，求$\angle AOB$的角平分线上的单位向量.
          \vspace{12em}

    \item 求圆锥面$2x^{2}+2y^{2}-z^{2}=0$与旋转抛物面$z=x^{2}+y^{2}$的交线在$P(1,
              1,2)$处的切线与法平面方程.
          \vspace{12em}

    \item 设$u=f(x,y,z)$具有连续的偏导数，$z=z(x,y)$是由方程$z^{5}-xz^{4}+yz^{3}=
              1$确定的隐函数，其中$5z^{2}-4xz+3y\neq0$，又$f'_{1}(0,0,1)=2$，$f'_{2}(0,0,
              1)=4$，$f'_{3}(0,0,1)=1$，求$\left.\mathrm{d}u\right|_{(0,0)}$.
          \vspace{12em}

    \item 设$z=yf(x^{2}-y^{2})$，其中$f$可导，求$\frac{1}{x}z_{x}+\frac{1}{y}z_{y}$.
          \vspace{12em}

    \item 求积分$I=\int_{1}^{3}\,\mathrm{d}x\int_{x-1}^{2}\sin y^{2}\,\mathrm{d}y$.
          \vspace{12em}

    \item 设平面区域$D=\{(x,y)|-2x\leq x^{2}+y^{2}\leq 4\}$，求二重积分$I=\iint\limits
              _{D}\left(y\cos x+\sqrt{x^{2}+y^{2}}\right)\,\mathrm{d}x\,\mathrm{d}y$.
          \vspace{12em}

    \item 设有直线$L:
              \begin{cases}
                  x+5y+z=0, \\
                  x-z+4=0
              \end{cases}$，平面$\pi:x-4y-8z+12=0$，求直线$L$在平面$\pi$上的投影直线的方程.
          \vspace{10em}

    \item 将空间曲线$\begin{cases}
                  x^{2}+y^{2}+z^{2}=1, \\
                  (x-1)^{2}+y^{2}+(z-1)^{2}=1
              \end{cases}$化为参数方程.
          \vspace{10em}

    \item 计算三重积分$I=\iiint\limits_{\Omega}\mathrm{e}^{1-z}\,\mathrm{d}x\,\mathrm{d}
              y\,\mathrm{d}z$，其中$\Omega$是三坐标面与平面$x+y+z=1$所围成的四面体.
          \vspace{10em}
\end{enumerate}

\section{综合题(每小题 8 分，共 40 分)}
\begin{enumerate}
    \setcounter{enumi}{10}
    \item 若函数$u=az^{4}-bxz+x^{2}+y^{2}$在点$P(1,1,1)$处沿方向$\boldsymbol
              {l}=(2,1,2)$的方向导数最大，求$a,b$的值，并求出最大方向导数.
          \vspace{13em}

    \item 已知函数$f(x,y)=x^{3}+y^{3}-(x+y)^{2}+2$，$D$是由$x+y=3$，$x=0$，$y
              =0$所围成的平面区域，求$f(x,y)$在$D$上的最大值与最小值.
          \vspace{13em}

    \item 设$z(x,y)=x^{y}+\int_{0}^{x}x\mathrm{e}^{-(y+t)^2}\,\mathrm{d}t$，求$\left
              .\frac{\partial^{2} z}{\partial x \partial y}\right|_{(1,0)}$.
          \vspace{12em}

    \item 设$f(x,y)=
              \begin{cases}
                  \frac{x^3y}{x^4+y^2}, & x^{4}+y^{2}\neq0, \\
                  0,                    & x^{4}+y^{2}=0.
              \end{cases}$讨论函数$f(x,y)$在原点处的连续性、偏导数存在性及可微性.
          \vspace{12em}

    \item 设函数$f(x),g(x)$在$[0,b]$上连续且单调增加，其中$b>0$，用\textbf{二重积分}证明:
          \[
              b\int_{0}^{b}f(x)g(x)\,\mathrm{d}x\geq \int_{0}^{b}f(x)\,\mathrm{d}x\int_{0}^{b}
              g(x)\,\mathrm{d}x
          \]
          \vspace{10em}
\end{enumerate}

\chapter{2023-2024学年微积分（一）（下）期中考试参考答案}
\section{基本计算题(每小题 6 分，共 60 分)}
\begin{enumerate}
    \item \textbf{Solution}. 由题意可得
          \[
              \begin{aligned}
                  0 = \boldsymbol{a}\cdot \left(\boldsymbol{a}\times (\boldsymbol{b}\times\boldsymbol{a})\right) = {} & \boldsymbol{a}\cdot (\boldsymbol{b}-2\boldsymbol{a})                                         \\
                  = {}                                                                                                & \boldsymbol{a}\cdot\boldsymbol{b}-2|\boldsymbol{a}|^{2}=\boldsymbol{a}\cdot\boldsymbol{b}-2.
              \end{aligned}
          \]
          所以$\boldsymbol{a}\cdot\boldsymbol{b}=2$.

          $|\boldsymbol{b}+\boldsymbol{a}| = \sqrt{(\boldsymbol{b}+\boldsymbol{a})^{2}}
              = \sqrt{|\boldsymbol{b}|^{2}+|\boldsymbol{a}|^{2}+2\boldsymbol{a}\cdot\boldsymbol{b}}
              = \sqrt{16+1+4}= \sqrt{21}$.

    \item \textbf{Solution}. 设$\alpha,\beta,\gamma$分别为$\overrightarrow{OA}$与$x$轴、$y$轴、$z$轴的夹角，

          则由题意可得$\cos \alpha = \frac{1}{2}$，$\cos \beta = \frac{\sqrt{2}}{2}$.
          因$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$，所以$\cos \gamma
              = \pm \frac{1}{2}$，

          又$\overrightarrow{OA}$在$z$轴上的坐标是负的，所以$\cos \gamma = -\frac{1}{2}$.
          从而$\overrightarrow{OA}= \left(\frac{1}{2},\frac{\sqrt{2}}{2},-\frac{1}{2}
              \right)$.

          $\overrightarrow{OB^0}=\left(\frac{1}{2},-\frac{\sqrt{2}}{2},-\frac{1}{2}\right
              )$，所以$\frac{\overrightarrow{OA}+\overrightarrow{OB^0}}{|\overrightarrow{OA}+\overrightarrow{OB^0}|}
              = \left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2}\right)$.

          即所求的$\angle AOB$的角平分线上的单位向量为$\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2}
              \right)$.

    \item \textbf{Solution}. 设$F(x,y,z) = 2x^{2}+2y^{2}-z^{2}$，$G(x,y,z)=x^{2}+
              y^{2}-z$.

          \textbf{法一} . 因
          \[
              \frac{\partial (F,G)}{\partial (y,z)}\bigg|_{P}=
              \begin{vmatrix}
                  4y & -2z \\
                  2y & -1
              \end{vmatrix}_{P}=\left.-4y+2yz\right|_{P}=4\neq 0,
          \]
          所以方程组$\begin{cases}
                  F(x,y,z)=0, \\
                  G(x,y,z)=0
              \end{cases}$可以唯一确定两个具有连续导数的隐函数$y=y(x),z=z(x)$.

          方程组$\begin{cases}
                  F(x,y,z) = 2x^{2}+2y^{2}-z^{2}=0, \\
                  G(x,y,z)=x^{2}+y^{2}-z=0
              \end{cases}$两边对$x$求导，得
          \[
              \begin{cases}
                  4x+4y\frac{\mathrm{d}y}{\mathrm{d}x}-2z\frac{\mathrm{d}z}{\mathrm{d}x}= 0, \\[6pt]
                  2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{\mathrm{d}z}{\mathrm{d}x}= 0.
              \end{cases}
          \]
          解得$\begin{cases}
                  \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{x}{y}, \\
                  \frac{\mathrm{d}z}{\mathrm{d}x} = 0
              \end{cases}$，所以切线的方向向量可取为$\left(1,\frac{\mathrm{d}y}{\mathrm{d}x}
              ,\frac{\mathrm{d}z}{\mathrm{d}x}\right)_{P}=(1,-1,0)$，

          切线的方程为$\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-2}{0}$，

          法平面的方程为$1\cdot(x-1)-1\cdot(y-1)+0\cdot(z-2)=0$，即$x-y=0$.

          \textbf{法二}. $\nabla F = (4x,4y,-2z)$，$\nabla G = (2x,2y,-1)$，

          取两个曲面的法向量分别为$\boldsymbol{n}_{F} = \frac{1}{4}\nabla F|_{P} = (1
              ,1,-1)$，$\boldsymbol{n}_{G} = \nabla G|_{P} = (2,2,-1)$，

          所以切线的方向向量可取为$\boldsymbol{n}_{F}\times \boldsymbol{n}_{G} =
              \begin{vmatrix}
                  \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\
                  1              & 1              & -1             \\
                  2              & 2              & -1
              \end{vmatrix}
              = (1,-1,0)$，

          切线的方程为$\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-2}{0}$，

          法平面的方程为$1\cdot(x-1)-1\cdot(y-1)+0\cdot(z-2)=0$，即$x-y=0$.

    \item \textbf{Solution}. 方程$z^{5}-xz^{4}+yz^{3}=1$两边全微分，得
          \[
              5z^{4}\,\mathrm{d}z-z^{4}\,\mathrm{d}x-4xz^{3}\,\mathrm{d}z+z^{3}\,\mathrm{d}y+3y
              z^{2}\,\mathrm{d}z=0,
          \]
          将$x=0,y=0,z=1$代入上式，得$5\,\mathrm{d}z-\mathrm{d}x+\mathrm{d}y=0$.所以
          \[
              \begin{aligned}
                  \left.\mathrm{d}u\right|_{(0,0)}= {} & f'_{1}\,\mathrm{d}x+f'_{2}\,\mathrm{d}y+f'_{3}\,\mathrm{d}z                   \\
                  = {}                                 & 2\,\mathrm{d}x+4\,\mathrm{d}y+\mathrm{d}z                                     \\
                  = {}                                 & 2\,\mathrm{d}x+4\,\mathrm{d}y+\frac{1}{5}\left(\mathrm{d}x-\mathrm{d}y\right) \\
                  = {}                                 & \frac{11}{5}\,\mathrm{d}x+\frac{19}{5}\,\mathrm{d}y.
              \end{aligned}
          \]

    \item \textbf{Solution}. 因为
          \[
              z_{x} = yf'(x^{2}-y^{2})\cdot 2x,\quad z_{y} = f(x^{2}
              -y^{2})+yf'(x^{2}-y^{2})\cdot (-2y),
          \]
          所以
          \[
              \frac{1}{x}z_{x}+\frac{1}{y}z_{y} = 2yf'(x^{2}-y^{2})+\frac{f(x^{2}-y^{2})}{y}
              -2yf'(x^{2}-y^{2}) = \frac{f(x^{2}-y^{2})}{y}.
          \]

    \item \textbf{Solution}.

          \noindent
          \begin{minipage}[c]{0.55\textwidth}
              积分区域$D$如右图所示.交换积分次序得
              \[
                  \begin{aligned}
                      I = {} & \int_{1}^{3}\,\mathrm{d}x\int_{x-1}^{2}\sin y^{2}\,\mathrm{d}y = \int_{0}^{2}\,\mathrm{d}y\int_{1}^{y+1}\sin y^{2}\,\mathrm{d}x \\
                      = {}   & \int_{0}^{2}y\sin y^{2}\,\mathrm{d}y = \left.-\frac{1}{2}\cos y^{2}\right|_{0}^{2}                                              \\
                      = {}   & -\frac{1}{2}\cos 4 + \frac{1}{2}= \frac{1}{2}(1-\cos 4).
                  \end{aligned}
              \]
          \end{minipage}
          \hfill %
          \begin{minipage}[c]{0.37\textwidth}
              \includegraphics[width=6cm]{2024x-figure1.pdf} % 引入图片源
          \end{minipage}

    \item \textbf{Solution}. 由于积分区域 $D$ 关于 $x$ 轴对称，所以 $\iint\limits
              _{D}y\cos x\,\mathrm{d}x\,\mathrm{d}y=0$.

          记 $D'=\{(x,y)|-2x\leq x^{2}+y^{2}\leq 4, y\geq 0\}$，则
          $I = 2\iint\limits_{D'}\sqrt{x^{2}+y^{2}}\,\mathrm{d}x\,\mathrm{d}y$.

          用极坐标代换，则

          \noindent
          \begin{minipage}[c]{0.5\textwidth}
              \[
                  \begin{aligned}
                      I & = 2\iint\limits_{D'}\sqrt{x^2+y^2}\,\mathrm{d}x\,\mathrm{d}y                                                                                                                   \\
                        & = 2\left(\int_{0}^{\frac{\pi}{2}}\,\mathrm{d}\theta\int_{0}^{2}r^{2}\,\mathrm{d}r+\int_{\frac{\pi}{2}}^{\pi}\,\mathrm{d}\theta\int_{-2\cos\theta}^{2}r^{2}\,\mathrm{d}r\right) \\
                        & = \frac{8}{3}\pi + \frac{16}{3}\int_{\frac{\pi}{2}}^{\pi}\left(1+\cos^{3}\theta\right)\,\mathrm{d}\theta                                                                       \\
                        & = \frac{16}{3}\pi + \frac{16}{3}\int_{\frac{\pi}{2}}^{\pi}\cos^{3}\theta\,\mathrm{d}\theta                                                                                     \\
                        & = \frac{16}{9}(3\pi-2).
                  \end{aligned}
              \]
          \end{minipage}
          \hfill %
          \begin{minipage}[c]{0.5\textwidth}
              \includegraphics[width=7cm]{2024x-figure2.pdf} % 引入图片源
          \end{minipage}

    \item \textbf{Solution}. 设过直线$L:
              \begin{cases}
                  x+5y+z=0, \\
                  x-z+4=0
              \end{cases}$的平面束方程为$x+5y+z+\lambda(x-z+4)=0$，

          即$(1+\lambda)x+5y+(1-\lambda)z+4\lambda=0$.

          令其与平面$\pi:x-4y-8z+12=0$垂直，得
          \[
              (1+\lambda)\cdot 1+5\cdot(-4)+(1-\lambda
              )\cdot(-8) = 0,
          \]
          解得$\lambda=3$.所以直线$L$在平面$\pi$上的投影直线的方程为
          \[
              \begin{cases}
                  4x+5y-2z+12=0, \\
                  x-4y-8z+12=0.
              \end{cases}
          \]

    \item \textbf{Solution}. 在方程组$\begin{cases}
                  x^{2}+y^{2}+z^{2}=1, \\
                  (x-1)^{2}+y^{2}+(z-1)^{2}=1
              \end{cases}$中消去$y$，得$(x-1)^{2}+(z-1)^{2}=x^{2}+z^{2}$，

          即$x+z=1$.将$z=1-x$代入$x^{2}+y^{2}+z^{2}=1$，得曲线在$xOy$平面上的投影柱面方程：$2
              x^{2}-2x+y^{2}=0$，

          即$4\left(x-\frac{1}{2}\right)^{2}+2y^{2}=1$. 令$\begin{cases}
                  x-\frac{1}{2} = \frac{1}{2}\cos \theta, \\[4.5pt]
                  y = \frac{1}{\sqrt{2}}\sin \theta,
              \end{cases}(0\leq \theta < 2\pi)$，则$z=1-x=\frac{1}{2}-\frac{1}{2}\cos\theta$，

          故曲线的参数方程为
          \[
              \begin{cases}
                  x = \frac{1}{2}+\frac{1}{2}\cos \theta, \\[6pt]
                  y = \frac{1}{\sqrt{2}}\sin \theta,      \\[6pt]
                  z = \frac{1}{2}-\frac{1}{2}\cos \theta.
              \end{cases}(0\leq \theta < 2\pi)
          \]

    \item \textbf{Solution}. 用截面法，
          \[
              \begin{aligned}
                  I = \iiint\limits_{\Omega}\mathrm{e}^{1-z}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = {} & \int_{0}^{1}\mathrm{e}^{1-z}\,\mathrm{d}z\iint\limits_{D_z}\,\mathrm{d}x\,\mathrm{d}y \qquad \left(D_{z}=\{(x,y)|x\geq 0,y\geq 0,x+y\leq 1-z\}\right)                           \\
                  = {}                                                                                   & \frac{1}{2}\int_{0}^{1}(1-z)^{2}\mathrm{e}^{1-z}\,\mathrm{d}z = \frac{1}{2}\int_{0}^{1}z^{2}\mathrm{e}^{z}\,\mathrm{d}z                                                         \\
                  = {}                                                                                   & \frac{1}{2}\int_{0}^{1}z^{2} \mathrm{d}\mathrm{e}^{z}= \frac{1}{2}\left(z^{2}\mathrm{e}^{z}\Big|_{0}^{1}-\int_{0}^{1}2z\mathrm{e}^{z}\,\mathrm{d}z\right)                       \\
                  = {}                                                                                   & \frac{\mathrm{e}}{2}-\int_{0}^{1}z\mathrm{e}^{z}\,\mathrm{d}z = \frac{\mathrm{e}}{2}-\int_{0}^{1}z\,\mathrm{d}\mathrm{e}^{z}                                                    \\
                  = {}                                                                                   & \frac{\mathrm{e}}{2}-\left(z\mathrm{e}^{z}\Big|_{0}^{1}-\int_{0}^{1}\mathrm{e}^{z}\,\mathrm{d}z\right) = \frac{\mathrm{e}}{2}-\left(\mathrm{e}-\left(\mathrm{e}-1\right)\right) \\
                  = {}                                                                                   & \frac{1}{2}\mathrm{e}-1.
              \end{aligned}
          \]

\end{enumerate}

\section{综合题(每小题 8 分，共 40 分)}
\begin{enumerate}
    \setcounter{enumi}{10}
    \item \textbf{Solution}. $\frac{\partial u}{\partial x}= -bz+2x$，$\frac{\partial
                  u}{\partial y}= 2y$，$\frac{\partial u}{\partial z}= 4az^{3}-bx$.

          所以$\nabla u|_{P}= \left.(-bz+2x,2y,4az^{3}-bx)\right|_{(1,1,1)}= (-b+2,2,
              4a-b)$.

          由于函数沿梯度方向的方向导数最大，所以$\nabla u|_{P}//\boldsymbol{l}$，有
          \[
              \frac{-b+2}{2}= \frac{2}{1}= \frac{4a-b}{2},
          \]
          解得$a=\frac{1}{2}$，$b=-2$. 此时$\nabla u|_{P}= (4,2,4)$，所以最大方向导数为$|
              \nabla u|_{P}| = \sqrt{4^{2}+2^{2}+4^{2}}= 6$.

    \item \textbf{Solution}. 先考虑$D$的内部. 令$f'_{x}= 3x^{2}-2(x+y) = f'_{y}
              = 3y^{2}-2(x+y) = 0$，

          解得唯一驻点$\left(\frac{4}{3},\frac{4}{3}\right)$，且
          \[
              f\left(\frac{4}{3},\frac{4}{3}
              \right)=\left(\frac{4}{3}\right)^{3}+\left(\frac{4}{3}\right)^{3}-\left(\frac{4}{3}
              +\frac{4}{3}\right)^{2}+ 2=-\frac{10}{27}.
          \]
          考虑$D$的坐标轴边界. 由对称性不妨只考虑在$y$轴上时，

          此时$x=0$，$f(x,0) = x^{3}-x^{2}+2$，令$\left(x^{3}-x^{2}+2\right)'=3x^{2}-
              2x=0$，解得$x=0$或$\frac{2}{3}$，且
          \[
              f\left(0,0\right) = 2,\quad f\left(\frac{2}{3},0\right) = f\left(0,\frac{2}{3}
              \right) = \frac{50}{27}.
          \]
          再考虑$D$的斜边边界，即直线$x+y=3$，

          此时$f = x^{3}+(3-x)^{3}-7$，令$\left(x^{3}+(3-x)^{3}-7\right)' = 3x^{2}-3(
                  3-x)^{2}=0$，解得$x=\frac{3}{2}$，且
          \[
              f\left(\frac{3}{2},\frac{3}{2}\right) = -\frac{1}{4},\quad f(3,0)=f(0,3) = 20.
          \]
          比较得函数$f(x,y)$在$D$上的最大值为$20$，最小值为$-\frac{10}{27}$.

    \item \textbf{Solution}. 令$u=y+t$，则$\mathrm{d}u = \mathrm{d}t$，$\int
              _{0}^{x}x\mathrm{e}^{-(y+t)^2}\,\mathrm{d}t = x\int_{y}^{y+x}\mathrm{e}^{-u^2}
              \mathrm{d}u$.

          所以
          \[
              z(x,y) = x^{y}+x\int_{y}^{y+x}\mathrm{e}^{-u^2}\,\mathrm{d}u,
          \]
          \[
              \frac{\partial z}{\partial x}= yx^{y-1}+\int_{y}^{y+x}\mathrm{e}^{-u^2}\,\mathrm{d}
              u+x\cdot \mathrm{e}^{-(y+x)^2},
          \]
          \[
              \begin{aligned}
                  \frac{\partial^2 z}{\partial x \partial y}= {} & \frac{\partial}{\partial y}\left(yx^{y-1}+\int_{y}^{y+x}\mathrm{e}^{-u^2}\,\mathrm{d}u+x\cdot \mathrm{e}^{-(y+x)^2}\right) \\
                  = {}                                           & x^{y-1}+yx^{y-1}\ln x + \mathrm{e}^{-(y+x)^2}-\mathrm{e}^{-y^2}+x\cdot \mathrm{e}^{-(y+x)^2}\cdot (-2)(y+x)                \\
                  = {}                                           & x^{y-1}\left(1+y\ln x\right)+\mathrm{e}^{-(y+x)^2}\left(1-2x(y+x)\right)-\mathrm{e}^{-y^2}.
              \end{aligned}
          \]

          从而$\left.\frac{\partial^{2} z}{\partial x \partial y}\right|_{(1,0)}= -\frac{1}{\mathrm{e}}$.

    \item \textbf{Solution}. 利用$\left(x^{2}-y^{2}\right)^{2}\geq 0$可得$\left
              |\frac{x^{3} y}{x^{4} + y^{2}}\right| \leq \frac{|x|}{2}$. 所以当$(x,y)\to(
              0,0)$时，$\left|\frac{x^{3} y}{x^{4} + y^{2}}\right| \leq \frac{|x|}{2}\to0$，

          由夹逼定理可知$\lim_{\substack{x\to 0 \\ y\to 0}}\frac{x^{3} y}{x^{4} + y^{2}}
              =0=f(0,0)$，因此函数$f(x,y)$在原点处连续.

          又$f(x,0)=f(0,y)\equiv 0$，所以$f_{x}(0,0)=f_{y}(0,0)=0$，因此函数$f(x,y)$在原点处的偏导数存在.

          考察
          \[
              \lim_{\substack{x\to 0 \\ y\to 0}}\frac{f(x,y)-f(0,0)-f_{x}(0,0)x-f_{y}(0,0)y}{\sqrt{x^{2}+y^{2}}}
              = \lim_{\substack{x\to 0 \\ y\to 0}}\frac{x^{3} y}{\left(x^{4} + y^{2}\right)\sqrt{x^2+y^2}}.
          \]
          当$y=x^{2},x\to0,y\to0$时，
          \[
              \lim_{\substack{x\to 0 \\ y\to 0}}\frac{x^{5}}{2x^{4}\sqrt{x^2+x^4}}
              = \lim_{x\to0}\frac{x}{2|x|\cdot\sqrt{x^2+1}}= \frac{1}{2}\lim_{x\to0}\frac{x}{|x|}
          \]
          不存在，所以函数$f(x,y)$在原点处不可微.

    \item \textbf{Solution}. 记区域$D=\{(x,y)|0\leq x\leq b,0\leq y\leq b\}$，$I
              =b\int_{0}^{b}f(x)g(x)\,\mathrm{d}x-\int_{0}^{b}f(x)\,\mathrm{d}x\int_{0}^{b}g(
              x)\,\mathrm{d}x$，则
          \[
              \begin{aligned}
                  I = {} & \int_{0}^{b}\,\mathrm{d}y\int_{0}^{b}f(x)g(x)\,\mathrm{d}x - \int_{0}^{b}f(x)\,\mathrm{d}x\int_{0}^{b}g(y)\,\mathrm{d}y \\
                  = {}   & \iint\limits_{D}f(x)g(x)\,\mathrm{d}x\,\mathrm{d}y - \iint\limits_{D}f(x)g(y)\,\mathrm{d}x\,\mathrm{d}y                 \\
                  = {}   & \iint\limits_{D}f(x)\left(g(x)-g(y)\right)\,\mathrm{d}x\,\mathrm{d}y.
              \end{aligned}
          \]

          注意到区域$D$关于直线$y=x$对称，具有轮换对称性，所以
          \[
              \begin{aligned}
                  I = {} & \iint\limits_{D}f(x)\left(g(x)-g(y)\right)\,\mathrm{d}x\,\mathrm{d}y                                                                                                \\
                  = {}   & \frac{1}{2}\left[\iint\limits_{D}f(x)\left(g(x)-g(y)\right)\,\mathrm{d}x\,\mathrm{d}y + \iint\limits_{D}f(y)\left(g(y)-g(x)\right)\,\mathrm{d}x\,\mathrm{d}y\right] \\
                  = {}   & \frac{1}{2}\iint\limits_{D}\left(f(x)-f(y)\right)\left(g(x)-g(y)\right)\,\mathrm{d}x\,\mathrm{d}y.
              \end{aligned}
          \]
          由于函数$f(x),g(x)$在$[0,b]$上单调增加，所以$\left(f(x)-f(y)\right)\left(g(
              x)-g(y)\right)\geq 0$，因此$I\geq 0$，即
          \[
              b\int_{0}^{b}f(x)g(x)\,\mathrm{d}x\geq \int_{0}^{b}f(x)\,\mathrm{d}x\int_{0}^{b}
              g(x)\,\mathrm{d}x.
          \]

\end{enumerate}
\end{document}





