# 换个角度，柳暗花明：交换积分次序

## 一、题目

$$I = \int_{0}^{1} \mathrm{~d} x \int_{x^{2}}^{1} \frac{x y}{\sqrt{1+y^{3}}} \mathrm{~d} y = ?$$

## 二、解析

$$x \in(0,1) \quad \mathrm{d} x^{2} \leqslant y \leqslant 1$$

$$I=\int_{0}^{1} x \mathrm{d} x \int_{x^{2}}^{1} \frac{y}{\sqrt{1+y^{3}}} \mathrm{d} y$$

$$y = x^{2} \Rightarrow x=\sqrt{y}$$

$$I=\int_{0}^{1} \mathrm{d} y \int_{0}^{\sqrt{y}} \frac{x y}{\sqrt{1+y^{3}}} \mathrm{d} x \Rightarrow$$

$$I=\int_{0}^{1} \frac{y}{\sqrt{1+y^{3}}} \mathrm{d} y \int_{0}^{\sqrt{y}} x \mathrm{d} x \Rightarrow$$

$$I=\int_{0}^{1} \frac{y}{\sqrt{1+y^{3}}} \cdot\left(\left.\frac{1}{2} x^{2}\right|_{0} ^{\sqrt{y}}\right) \mathrm{d} y \Rightarrow$$

$$I=\frac{1}{2} \int_{0}^{1} \frac{y^{2}}{\sqrt{1+y^{3}}} \mathrm{d} y \Rightarrow\left(y^{3}\right)_{y}^{\prime}=3 y^{2} \Rightarrow$$

$${\left[\left(1+y^{3}\right)^{\frac{1}{2}}\right]_{y}^{\prime}=\frac{1}{2}\left(1+y^{3}\right)^{-\frac{1}{2}} \cdot 3 y^{2}} \Rightarrow$$

$$I=\frac{2}{3} \cdot \frac{1}{2}\left(\left.\sqrt{1+y^{3}}\right|_{0} ^{1}\right)=\frac{1}{3}(\sqrt{2}-1)$$

### 特别专题

Copyright © 2023   zhaokaifeng.com   All Rights Reserved.