怎么判断是否需要交换二重积分得积分次序？

一、题目

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

难度评级：

二、解析

首先，我们会发现，不交换积分次序十分不好算：

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

于是，我们只能尝试交换积分次序。

要交换积分次序，首先要绘制出原式得积分区域，如图 01 所示：

接着：

$$
\begin{aligned}
I & = \int_{0}^{1} \mathrm{~d} y \int_{0}^{\sqrt{y}} \frac{x y}{\sqrt{1+y^{3}}} \mathrm{~d} x \\ \\
& = \int_{0}^{1} \frac{y}{\sqrt{1+y^{3}}} \mathrm{~d} y \int_{0}^{\sqrt{y}} x \mathrm{~d} x \\ \\
& = \left.\int_{0}^{1} \frac{y}{\sqrt{1+y^{3}}} \cdot \frac{1}{2} x^{2}\right|_{0} ^{\sqrt{y}} \mathrm{~d} y \\ \\
& = \frac{1}{2} \int_{0}^{1} \frac{y}{\sqrt{1+y^{3}}} \cdot y \mathrm{~d} y= \\ \\
& = \frac{1}{2} \int_{0}^{1} \frac{y^{2}}{\sqrt{1+y^{3}}} \mathrm{~d} y \\ \\
& =\frac{1}{6} \int_{0}^{1} \frac{1}{\sqrt{1+y^{3}}} \mathrm{~d} \left(y^{3}\right)
\end{aligned}
$$

令 $t=y^{3}$, 则：

$$
t \in(0,1)
$$

于是：

$$
\begin{aligned}
I & = \frac{1}{6} \int_{0}^{1}(1+t)^{\frac{-1}{2}} \mathrm{~d} t \\ \\
& = \left.\frac{1}{6} \cdot 2 \cdot (1+t)^{\frac{1}{2}}\right|_{0} ^{1} \\ \\
& = \frac{2}{6}(\sqrt{2}-1)=\frac{\sqrt{2}-1}{3}
\end{aligned}
$$

注意：
$$
(\sqrt{1+t})_{t}^{\prime}=\frac{1}{2} \cdot \frac{1}{\sqrt{1+t}}
$$

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