# 化繁为简：以 $\int_{1}^{2}$ $(x-1)^{2}$ $(x-2)^{2}$ $\mathrm{d} t$ 为例

## 一、题目

$$\int_{1}^{2} (x-1)^{2} (x-2)^{2} \mathrm{d} t =$$

## 二、解析

### 方法一：换元法

$$t = x -1$$

$$x = t + 1$$

$$\mathrm{d} x = \mathrm{d} t$$

$$x \in (1, 2) \Rightarrow t \in (0, 1)$$

Next

$$\int_{1}^{2} (x-1)^{2} (x-2)^{2} \mathrm{d} t =$$

$$\int_{0}^{1} t^{2} (t-1)^{2} \mathrm{d} t =$$

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

$$\int_{0}^{1} (t^{4} + t^{2} – 2t^{3}) \mathrm{d} t =$$

$$\Big( \frac{1}{5} t^{5} + \frac{1}{3} t^{3} – 2 \cdot \frac{1}{4} t^{4} \Big) \Big|_{0}^{1} =$$

$$\frac{1}{5} + \frac{1}{3} – \frac{1}{2} = \frac{8}{15} – \frac{1}{2} = \frac{16}{30} – \frac{15}{30} = \frac{1}{30}$$

### 方法二：分部积分法

$$[(x-1)^{3}]^{\prime} = 3 (x-1)^{2}$$

Next

$$\int_{1}^{2} (x-1)^{2} (x-2)^{2} \mathrm{d} t =$$

$$\frac{1}{3} \int_{1}^{2} (x-2)^{2} \mathrm{d} [(x-1)^{3}] =$$

$$\frac{1}{3} (x-2)^{2}(x-1)^{3} \big|_{1}^{2} – \frac{1}{3} \int{1}^{2} (x-1)^{3} \mathrm{d} [(x-2)^{2}] =$$

$$0 – \frac{1}{3} \int_{1}^{2} (x-1)^{3} \mathrm{d} [(x-2)^{2}] =$$

$$– \frac{2}{3} \int_{1}^{2} (x-1)^{3} (x-2) \mathrm{d} x =$$

Next

$$– \frac{2}{3} \cdot \frac{1}{4} \int_{1}^{2} (x-2) \mathrm{d} [(x-1)^{4}] =$$

$$\frac{1}{6} (x-2) (x-1)^{4} \Big|_{1}^{2} + \frac{1}{6} \int{1}^{2} (x-1)^{4} \mathrm{d} (x-2) =$$

$$0 + \frac{1}{6} \int_{1}^{2} (x-1)^{4} \mathrm{d} x =$$

$$\frac{1}{6} \cdot \frac{1}{5} (x-1)^{5} \Big|_{1}^{2} = \frac{1}{6} \cdot \frac{1}{5} = \frac{1}{30}$$