重庆大学2004年数学分析330
四、解:
$a_n=\frac{1}{n}((3+(-1)^n)x)^n.$
$|R|=|\frac{\frac{1}{n}}{\frac{1}{n+1}}|=1,(n \to \infty )$
$\therefore -1<((3+(-1)^n)x)<1,$
$\Rightarrow -\frac{1}{4}<x<\frac{1}{4}.$
五、解:先作变量变换,
$u=xy,v=\frac{y^2}{x},$
$\therefore x=\sqrt[3]{\frac{u^2}{v}},y=\sqrt[3]{uv},$
$|J|=\begin{vmatrix}
\frac{2}{3}\frac{1}{\sqrt[3]{uv}} & -\frac{1}{3}\sqrt[3]{\frac{u^2}{v^4}}\\
\frac{1}{3}\sqrt[3]{\frac{v}{u^2}} & \frac{1}{3}\sqrt[3]{\frac{u}{v^2}}
\end{vmatrix}=\frac{2}{9v}+\frac{1}{9v}=\frac{1}{3v}.$
$D\rightarrow D':1\leq u\leq 3,1\leq v\leq 3.$
$\displaystyle I=\iint_D\frac{3x}{y^2+xy^3}dxdy=\iint_{D'}\frac{1}{v^2(1+u)}dudv=\int_{1}^{3}\frac{dv}{v^2}\int_{1}^{3}\frac{du}{1+u}=\frac{2}{3}\ln2.$