# 2022考研数二第02题解析：更改积分次序、定积分中的变量替换

## 二、解析

\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}

$$\int _{ 0 } ^ { 2 } \mathrm { ~ d } y \int _{ y } ^ { 2 } \frac { y } { \sqrt { 1 + x ^ { 3 } } } \mathrm { ~ d } x = \textcolor{springgreen}{ \int _{ 0 } ^ { 2 } \mathrm{~d} x \int _{ 0 } ^ { x } \frac { y } { \sqrt { 1 + x ^ { 3 } } } \mathrm{~d} y }$$

\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}

\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}