# 2018年考研数二第16题解析：变上限积分、一阶线性微分方程、积分中值定理

## 题目

$$\int_{0}^{x} f(t) \mathrm{d} t + \int_{0}^{x} t f(x – t) \mathrm{d} t = ax^{2}.$$

$(Ⅰ)$ 求 $f(x)$

$(Ⅱ)$ 若 $f(x)$ 在区间 $[0, 1]$ 上的平均值为 $1$, 求 $a$ 的值.

## 解析

### 第 $(Ⅰ)$ 问

$$\int_{0}^{x} t f(x – t) \mathrm{d} t \Rightarrow$$

$$\left\{\begin{matrix} u = x – t;\\ t = x – u. \end{matrix}\right. \Rightarrow$$

$$\int_{x}^{0} (x – u) f(u) \mathrm{d} (x – u) \Rightarrow$$

[1]. 当 $t \in (0, x)$ 时，$x – t \in (x, 0)$, 即 $u \in (x, 0)$.

$$(-1) \cdot \int_{x}^{0} (x – u) f(u) \mathrm{d} u \Rightarrow$$

$$\int_{0}^{x} (x – u) f(u) \mathrm{d} u \Rightarrow$$

$${\color{Red} x \int_{0}^{x} f(u) \mathrm{d} u – \int_{0}^{x} u f(u) \mathrm{d} u}.$$

$$\int_{0}^{x} f(t) \mathrm{d} t + \int_{0}^{x} t f(x – t) \mathrm{d} t = ax^{2} \Rightarrow$$

$$\int_{0}^{x} f(t) \mathrm{d} t + x \int_{0}^{x} f(u) \mathrm{d} u – \int_{0}^{x} u f(u) \mathrm{d} u = ax^{2} \Rightarrow$$

$$等号两端同时求导 \Rightarrow$$

$$f(x) + \int_{0}^{x} f(u) \mathrm{d} u + x f(x) – x f(x) = 2ax \Rightarrow$$

$$f(x) + \int_{0}^{x} f(u) \mathrm{d} u = 2ax \Rightarrow$$

$$等号两端再次同时求导 \Rightarrow$$

$$f^{‘}(x) + f(x) = 2a \Rightarrow$$

$$一阶线性微分方程求解公式 \Rightarrow$$

$$f(x) = \Bigg[ \int 2a e^{ \int 1 \mathrm{d} x} \mathrm{d} x + C \Bigg] \cdot e^{\int (-1) \mathrm{d} x} \Rightarrow$$

$$f(x) = \Bigg[ 2a \int e^{x} \mathrm{d} x + C \Bigg] \cdot e^{-x} \Rightarrow$$

$$f(x) = (2a e^{x} + C) \cdot e^{-x} \Rightarrow$$

$$f(x) = 2a + Ce^{-x}.$$

$$\int_{0}^{x} f(t) \mathrm{d} t + \int_{0}^{x} t f(x – t) \mathrm{d} t = ax^{2} \Rightarrow$$

$$\int_{0}^{0} f(t) \mathrm{d} t + \int_{0}^{0} t f(x – t) \mathrm{d} t = 0 \Rightarrow$$

$$0 = 0 \Rightarrow$$

$$f(0) = 0.$$

$$f(0) = 2a + C = 0 \Rightarrow$$

$$C = -2a.$$

$$f(x) = 2a – 2a e^{-x} \Rightarrow$$

$$f(x) = 2a (1-e^{-x}).$$

### 第 $(Ⅱ)$ 问

$$\frac{\int _{0} ^{1} 2a (1-e^{-x}) \mathrm{d} x}{1 – 0} = 1 \Rightarrow$$

$$\int _{0} ^{1} 2a (1-e^{-x}) \mathrm{d} x = 1 \Rightarrow$$

$$2a \int _{0} ^{1} \mathrm{d} x – 2a \int_{0}^{1} e^{-x} \mathrm{d} x = 1 \Rightarrow$$

$$2a – 2a \cdot(- e^{-x} |_{0}^{1}) = 1 \Rightarrow$$

$$2a – 2a(\frac{-1}{e} + 1) \Rightarrow$$

$$2a – 2a(1 – \frac{1}{e}) \Rightarrow$$

$$2a – 2a + \frac{2a}{e} = 1 \Rightarrow$$

$$\frac{2a}{e} = 1 \Rightarrow$$

$$2a = e \Rightarrow$$

$$a = \frac{e}{2}.$$