用三角代换、几何意义和区间再现三种方法解一道定积分题目

一、题目

$$
\int_{0}^{2} x \sqrt{2x – x^{2}} \mathrm{d} x = ?
$$

本文的题目解析中提供了三种不同角度的解法。

难度评级：

二、解析

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方法一：三角代换

由于 $(1-x)^{2}$ $=$ $1$ $+$ $x^{2}$ $-$ $2x$.

因此，$1$ $-$ $(1-x)^{2}$ $=$ $1$ $-$ $1$ $-$ $x^{2}$ $+$ $2x$ $=$ $2x$ $-$ $x^{2}$.

于是，可令：

$$
\sin t = 1 – x
$$

进而有：

$$
x \in (0, 2) \Rightarrow
$$

$$
\sin t \in (1-0, 1-2) \Rightarrow \sin t \in (1, -1) \Rightarrow t \in (\frac{\pi}{2}, -\frac{\pi}{2})
$$

$$
x = 1 – \sin t \Rightarrow \mathrm{d} x = – \cos t \mathrm{d} t.
$$

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于是：

$$
\int_{0}^{2} x \sqrt{2x – x^{2}} \mathrm{d} x =
$$

$$
\int_{0}^{2} x \sqrt{1 – (1 – x)^{2}} \mathrm{d} x =
$$

$$
– \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} (1 – \sin t) \cos^{2} t \mathrm{d} t =
$$

$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2} t – \sin t \cdot \cos^{2} t) \mathrm{d} t =
$$

$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2} t \mathrm{d} t – \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin t \cdot \cos^{2} t \mathrm{d} t =
$$

$$
2 \int_{0}^{\frac{\pi}{2}} \cos^{2} t \mathrm{d} t – 0 = 2 \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{2}.
$$

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方法二：几何意义

$$
\int_{0}^{2} x \sqrt{2x – x^{2}} \mathrm{d} x =
$$

$$
\int_{0}^{2} x \sqrt{1 – (x – 1)^{2}} \mathrm{d} x =
$$

$$
\int_{0}^{2} [(x-1) + 1] \sqrt{1 – (x – 1)^{2}} \mathrm{d} x =
$$

$$
\int_{0}^{2} (x-1) \sqrt{1 – (x – 1)^{2}} \mathrm{d} x + \int_{0}^{2} \sqrt{1 – (x – 1)^{2}} \mathrm{d} x =
$$

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其中，$(x-1) \sqrt{1 – (x – 1)^{2}}$ 是关于 $x = 1$ 反对称的奇函数，因此 $\int_{0}^{2}$ $(x-1) \sqrt{1 – (x – 1)^{2}}$ $\mathrm{d} x$ $=$ $0$ ——关于此的详细阐述，请查看《关于非原点位置反对称的“奇函数”》这篇文章。

$$
\int_{0}^{2} \sqrt{1 – (x – 1)^{2}} \mathrm{d} x = \frac{\pi}{2}.
$$

若令 $y$ $=$ $\sqrt{1 – (x – 1)^{2}}$, 则可得 $(x – 1)^{2}$ $+$ $y^{2}$ $=$ $1$——根据 圆形的标准方程 可知，这其实是一个圆心为 $(1, 0)$, 半径为 $1$ 的位于 $x$ 轴上方的上半圆，该上半圆的面积为 $\frac{\pi}{2}$.

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方法三：区间再现

令 $t$ $=$ $2 + 0 – x$, 即：

$$
t = 2 -x
$$

$$
x = 2 – t
$$

$$
\mathrm{d} x = – \mathrm{d} t
$$

$$
x \in (0, 2) \Rightarrow t \in (2, 0)
$$

Next

从而：

$$
\int_{0}^{2} x \sqrt{2x – x^{2}} \mathrm{d} x =
$$

$$
\int_{2}^{0} (2 – t) \sqrt{2(2 – t) – (2 – t)^{2}} (- \mathrm{d} t) =
$$

$$
\int_{0}^{2} (2 – t) \sqrt{2(2 – t) – (2 – t)^{2}} \mathrm{d} t =
$$

$$
\int_{0}^{2} (2 – t) \sqrt{2 t – t^{2}} \mathrm{d} t =
$$

$$
\int_{0}^{2} 2 \sqrt{2 t – t^{2}} \mathrm{d} t – \int_{0}^{2} t \sqrt{2 t – t^{2}} \mathrm{d} t \Rightarrow
$$

Next

令 $t$ $=$ $x$ $\Rightarrow$

$$
\int_{0}^{2} 2 \sqrt{2 x – x^{2}} \mathrm{d} x – \int_{0}^{2} x \sqrt{2 x – x^{2}} \mathrm{d} x \Rightarrow
$$

$$
2 \int_{0}^{2} \sqrt{2 x – x^{2}} \mathrm{d} x – \int_{0}^{2} x \sqrt{2 x – x^{2}} \mathrm{d} x = \int_{0}^{2} x \sqrt{2 x – x^{2}} \mathrm{d} x \Rightarrow
$$

$$
2 \int_{0}^{2} \sqrt{2 x – x^{2}} \mathrm{d} x = 2 \int_{0}^{2} x \sqrt{2 x – x^{2}} \mathrm{d} x \Rightarrow
$$

$$
\int_{0}^{2} \sqrt{2 x – x^{2}} \mathrm{d} x = \int_{0}^{2} x \sqrt{2 x – x^{2}} \mathrm{d} x
$$

Next

又根据几何意义可知，$\int_{0}^{2}$ $\sqrt{2 x – x^{2}}$ $\mathrm{d} x$ $=$ $\int_{0}^{2}$ $\sqrt{1 – (x – 1)^{2}}$ $\mathrm{d} x$ $=$ $\frac{\pi}{2}$, 于是：

$$
\int_{0}^{2} x \sqrt{2 x – x^{2}} \mathrm{d} x = \frac{\pi}{2}
$$

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