一些数学分析考研题（2024年）的解析

解
      由于\[\begin{align*}\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)}x^{n}
&=\frac{1}{x}\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)}x^{n+1} \\
&=\frac{1}{x}\int\limits(\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)}x^{n+1})'dx \\
&=\frac{1}{x}\int\limits\displaystyle\sum_{n=1}^{\infty }x^{n}dx \\
&=\frac{1}{x}\int\limits\frac{x}{(1-x)}dx \\
&=\frac{1}{x}[x-\ln(1-x)] .
\end{align*}\]令$x=\frac{1}{2}$，代入上式，得\[\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)2^n}=1+2\ln2.\]

解    由于\[\sqrt{1+\tan x}\backsim 1+\frac{1}{2}\tan x=1+\frac{1}{2}\sin x,\]\[\sqrt{1-\sin x}\backsim 1-\frac{1}{2}\sin x,\]所以\[\lim_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1-\sin x}}{x}=\lim_{x\to0}\frac{1+\frac{1}{2}\sin x-1+\frac{1}{2}\sin x}{x}=1.\]

解
         由于\[\frac{\mathrm{d}z}{\mathrm{d}x}=2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}.\]\[2x+y+x\frac{\mathrm{d}y}{\mathrm{d}x}+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0,\]\[\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2x+y}{x+2y}.\]\[\therefore  \frac{\mathrm{d}z}{\mathrm{d}x}=2x-2y\frac{2x+y}{x+2y}=\frac{2x^2-2y^2}{x+2y}.\]\[\frac{\mathrm{d}^2z}{\mathrm{d} x^2}=(\frac{2x^2-2y^2}{x+2y})'=\cdots .\]

解  由\[\begin{cases}
y^2=2x\\
x^2+y^2=8
\end{cases}\]\[解得两曲线的交点(2,2),(2,-2).设第一象限内两曲线共有区域的面积为\varDelta _1，\]\[圆被抛物线分割的另一部分面积为\varDelta _2.则\]\[\begin{align*}\varDelta _1
&=\int_{0}^{2}[\sqrt{8-y^2}-\frac{1}{2}y^2]dy \\
&=\int_{0}^{2}\sqrt{8-y^2}dy-\int_{0}^{2}\frac{1}{2}y^2dy \\
&=2+\pi-\frac{4}{3} =\frac{2}{3}+\pi .
\end{align*}\]\[左右面积之比=\frac{8\pi -2\varDelta _1}{2\varDelta _1}=\frac{6\pi -\frac{4}{3}}{2\pi +\frac{4}{3}}.\]


      

解  用参数方程\[x=a\cos \theta ,y=b\sin \theta .0\le \theta \le \frac{\pi }{2}.\]从而对弧长的曲线积分\[\begin{align*}I
&=\int_{L}xyds \\
&=ab\int_{0}^{\frac{\pi }{2}}\cos \theta \sin \theta\sqrt{a^2\sin ^2\theta+b^2\cos ^2\theta}d\theta  \\
&=ab\int_{0}^{\frac{\pi }{2}}\frac{\sqrt{a^2\sin ^2\theta+b^2\cos ^2\theta}d[(a^2-b^2)\sin^2 \theta+b^2 ]}{2(a^2-b^2)}\\
&=\frac{ab(a^2+ab+b^2)}{3(a+b)}.
\end{align*}\]

解 \[直线z=y,x=0绕z轴一圈的空间方程为：z^2=x^2+y^2.\]\[\Omega :z^2=x^2+y^2,z=1\]\[\begin{align*}I
&=\iiint\limits_{\Omega }(x^2+y^2+z)dxdydz \\
&=\iint\limits_{D} dxdy\int_{\sqrt{x^2+y^2}}^{1}(x^2+y^2+z)dz\\
&=\iint\limits_{D}(x^2+y^2)(1-\sqrt{x^2+y^2}) dxdy+\frac{1}{2}\iint\limits_{D}(1-(x^2+y^2))dxdy\\
&=\int_{0}^{2\pi }d\theta \int_{0}^{1}r^2(1-r^2)rdr+\frac{1}{2}\int_{0}^{2\pi }d\theta \int_{0}^{1}(1-r^2)rdr\\
&=\frac{\pi }{10} +\frac{\pi }{4}\\
&=\frac{7}{20}\pi .
\end{align*}\]

解 \[f(x)为偶函数，周期2\pi .b_n=0.(n=1,2,\cdots ).\]\[\begin{align*}a_0
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\cos x)dx \\
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\sin (\frac{\pi }{2}-x))dx \\
&=\frac{2}{\pi }\int_{0}^{\pi }(\frac{\pi }{2}-x)dx\\
&=0.
\end{align*}\]\[\begin{align*}a_n
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\cos x)\cos nx dx \\
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\sin (\frac{\pi }{2}-x))\cos nx dx \\
&=\frac{2}{\pi }\int_{0}^{\pi } (\frac{\pi }{2}-x)\cos nx dx \\
&=\frac{2}{n\pi }[ (\frac{\pi }{2}-x)\sin nx-\frac{1}{n} \cos nx]|_0^\pi  \\
&=\begin{cases}
0,& x=2k\\
\frac{4}{(2k-1)^2\pi },& n=2k-1
\end{cases}(k=1,2,\cdots )
\end{align*}\]
由狄氏收敛定理，得\[f(x)=\arcsin (\cos x)=\frac{4}{\pi }\displaystyle\sum_{n=1}^{\infty }\frac{1}{(2n-1)^2}\cos (2n-1)x.\]

解 由积分区域的对称性可知：\[\iint_{S}xdS=\iint_{S}ydS=0.\]所以\[\begin{align*}I &=\iint_{S}(x+y+z)dS= \iint_{S}zdS\\
&=\iint_{D}\sqrt{a^2-x^2-y^2}\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}}dxdy \\
&=\iint_{D}\sqrt{a^2-x^2-y^2}\cdotp \frac{a}{\sqrt{a^2-x^2-y^2}} dxdy\\
&=a\iint_{D}dxdy=a\cdotp \pi a^2=\pi a^3.
\end{align*}\]




   

证明  \[\because \lim_{n\to\infty }a_n=a.\]\[\therefore \forall \varepsilon > 0,\exists N\in \mathbb{N},n> N,s.t.|a_n-a|< \varepsilon .\]对于上面的$N$,取\[\varepsilon =1.此时有a-1< a_n< a+1,\]而\[\sqrt[n]{a-1}< \sqrt[n]{a_n}< \sqrt[n]{a+1}.\]又因为有\[\lim_{n\to\infty }\sqrt[n]{a}=1,\]故由夹逼法得到\[\lim_{n\to\infty }\sqrt[n]{a_n}=1。\]