# 求解定积分 $\int_{0}^{2}$ $x \sqrt{2x – x^{2}}$ $\mathrm{d} x$

## 一、题目

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

## 二、解析

Next

### 方法一：三角代换

$$\sin t = 1 – x$$

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

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

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

Next

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

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

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

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

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

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

Next

### 方法二：几何意义

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

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

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

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

Next

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

Next

### 方法三：区间再现

$$t = 2 -x$$

$$x = 2 – t$$

$$\mathrm{d} x = – \mathrm{d} t$$

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

Next

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

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

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

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

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

Next

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

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

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

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

Next

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