一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
$$
\int _{ 0 } ^ { 2 } \mathrm { ~ d } y \int _{ y } ^ { 2 } \frac { y } { \sqrt { 1 + x ^ { 3 } } } \mathrm{~d} x = ?
$$
A. $\frac { \sqrt { 2 } } { 6 }$
B. $\frac { 1 } { 3 }$
C. $\frac { \sqrt { 2 } } { 3 }$
D. $\frac{2}{3}$
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,根据 “$\int_{0}^{2} \mathrm{~d} x$” 可知,积分区域位于 $x = 0$ 到 $x = 2$ 之间;接着,根据 “$\int_{y}^{2} \mathrm{~d} y$” 可知,积分区域同时也位于 $x = y$ 到 $x = 2$ 之间。于是,积分区域 $D$ 如图 01 阴影部分所示:
观察可知,如果按照现在的积分次序,我们可以对原积分式子做出如下变形:
$$
\begin{aligned}
\int _{ 0 } ^ { 2 } \mathrm { ~ d } y \int _{ y } ^ { 2 } \frac { \textcolor{brown}{y} } { \textcolor{tan}{ \sqrt {1 + x ^ { 3 } }} } \mathrm{~d} x \\ \\
& = \int _{ 0 } ^ { 2 } \textcolor{brown}{y} \mathrm { ~ d } y \int _{ y } ^ { 2 } \textcolor{tan}{\frac { 1 } {\sqrt { 1 + x ^ { 3 }}} } \mathrm{~d} x \\ \\
\end{aligned}
$$
对上面式子中的 “$\int _{ y } ^ { 2 } \textcolor{tan}{\frac { 1 } {\sqrt { 1 + x ^ { 3 }}} } \mathrm{~d} x$” 进行积分相当困难——
因此,我们尝试更改积分次序,从先对 $x$ 积分后对 $y$ 积分,更改为先对 $y$ 积分,后对 $x$ 积分:
$$
\int _{ 0 } ^ { 2 } \mathrm { ~ d } y \int _{ y } ^ { 2 } \frac { y } { \sqrt { 1 + x ^ { 3 } } } \mathrm { ~ d } x = \textcolor{springgreen}{ \int _{ 0 } ^ { 2 } \mathrm{~d} x \int _{ 0 } ^ { x } \frac { y } { \sqrt { 1 + x ^ { 3 } } } \mathrm{~d} y }
$$
于是:
$$
\begin{aligned}
\textcolor{springgreen}{\int _{ 0 } ^ { 2 } \mathrm{~d} x \int _{ 0 } ^ { x } \frac { y } { \sqrt { 1 + x ^ { 3 } } } \mathrm{~d} y} \\ \\
& = \int _{ 0 } ^ { 2 } \frac{1}{\sqrt { 1 + x ^ { 3 } }} \mathrm{~d} x \int _{ 0 } ^ { x } y \mathrm{~d} y \\ \\
& = \int _{ 0 } ^ { 2 } \frac{1}{\sqrt { 1 + x ^ { 3 } }} \cdot \left( \frac{1}{2} y ^{2} \Big|_{0}^{x} \right) \mathrm{~d} x \\ \\
& = \int _{ 0 } ^ { 2 } \frac { x ^ { 2 } } { \textcolor{pink}{\boldsymbol{2}} \sqrt { 1 + x ^ { 3 } } } \mathrm{~d} x \\ \\
& = \textcolor{pink}{\boldsymbol{\frac{1}{2}}} \int _{ 0 } ^ { 2 } \frac { x ^ { 2 } } { \sqrt { 1 + x ^ { 3 } } } \mathrm{~d} x \\ \\
& = \textcolor{pink}{\boldsymbol{ \frac{1}{2} }} \cdot \frac{1}{3} \cdot \textcolor{yellow}{\boldsymbol{ \int _{ 0 } ^ { 2 } \frac { 1 } { \sqrt { x ^ { 3 } + 1 } } \mathrm{~d} \left( x ^ { 3 } + 1 \right) }} \\ \\
& \xlongequal[t \in \left( 1, 9 \right)]{t=x ^{3} + 1} \frac{1}{6} \int_{\textcolor{orangered}{1}}^{9} \frac{1}{\sqrt{t}} \mathrm{~d} t \\ \\
& = \frac{1}{6} \cdot 2 \cdot \sqrt{t} \Big|_{\textcolor{orangered}{1}} ^{9} \\ \\
& = \frac{1}{3} \cdot \left( 3-1 \right) \\ \\
& = \frac { 2 } { 3 }
\end{aligned}
$$
当然,对上面的 “$\textcolor{yellow}{\boldsymbol{ \int _{ 0 } ^ { 2 } \frac { 1 } { \sqrt { x ^ { 3 } + 1 } } \mathrm{~d} \left( x ^ { 3 } + 1 \right) }}$” 这个式子还可以有下面的计算方式:
$$
\begin{aligned}
\textcolor{yellow}{\boldsymbol{ \int _{ 0 } ^ { 2 } \frac { 1 } { \sqrt { x ^ { 3 } + 1 } } \mathrm{~d} \left( x ^ { 3 } + 1 \right) }} \\ \\
& = \int _{ 0 } ^ { 2 } \frac { 1 } { \sqrt { x ^ { 3 } + 1 } } \mathrm{~d} \left( x ^ { 3 }\right) \\ \\
& \xlongequal[t \in \left( 0, 8 \right)]{t = x ^{3}} \int_{\textcolor{orangered}{0}}^{8} \frac{1}{\sqrt{t+1}} \mathrm{~d} t \\ \\
& = 2 \cdot \sqrt{t+1} \Big|_{\textcolor{orangered}{0}}^{8} \\ \\
& = 2 \left( 3-1 \right) \\ \\
& = 4
\end{aligned}
$$
综上可知,本 题 应 选 D