题目
已知连续函数 $f(x)$ 满足：
$$
\int_{0}^{x} f(t) \mathrm{d} t + \int_{0}^{x} t f(x – t) \mathrm{d} t = ax^{2}.
$$
$(Ⅰ)$ 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
首先，由题可知：
$$
\int_{0}^{x} t f(x – t) \mathrm{d} t \Rightarrow
$$
$$
令
\left\{\begin{matrix}
u = x – t;\\
t = x – u.
\end{matrix}\right.
\Rightarrow
$$
$$
\int_{x}^{0} (x – u) f(u) \mathrm{d} (x – u) \Rightarrow
$$
注：
[1]. 当 $t \in (0, x)$ 时，$x – t \in (x, 0)$, 即 $u \in (x, 0)$.
$$
(-1) \cdot \int_{x}^{0} (x – u) f(u) \mathrm{d} u \Rightarrow
$$
$$
\int_{0}^{x} (x – u) f(u) \mathrm{d} u \Rightarrow
$$
$$
{\color{Red}
x \int_{0}^{x} f(u) \mathrm{d} u – \int_{0}^{x} u f(u) \mathrm{d} u}.
$$
于是：
$$
\int_{0}^{x} f(t) \mathrm{d} t + \int_{0}^{x} t f(x – t) \mathrm{d} t = ax^{2} \Rightarrow
$$
$$
\int_{0}^{x} f(t) \mathrm{d} t + x \int_{0}^{x} f(u) \mathrm{d} u – \int_{0}^{x} u f(u) \mathrm{d} u = ax^{2} \Rightarrow
$$
$$
等号两端同时求导 \Rightarrow
$$
$$
f(x) + \int_{0}^{x} f(u) \mathrm{d} u + x f(x) – x f(x) = 2ax \Rightarrow
$$
$$
f(x) + \int_{0}^{x} f(u) \mathrm{d} u = 2ax \Rightarrow
$$
$$
等号两端再次同时求导 \Rightarrow
$$
$$
f^{‘}(x) + f(x) = 2a \Rightarrow
$$
$$
一阶线性微分方程求解公式 \Rightarrow
$$
$$
f(x) = \Bigg[ \int 2a e^{ \int 1 \mathrm{d} x} \mathrm{d} x + C \Bigg] \cdot e^{\int (-1) \mathrm{d} x} \Rightarrow
$$
$$
f(x) = \Bigg[ 2a \int e^{x} \mathrm{d} x + C \Bigg] \cdot e^{-x} \Rightarrow
$$
$$
f(x) = (2a e^{x} + C) \cdot e^{-x} \Rightarrow
$$
$$
f(x) = 2a + Ce^{-x}.
$$
又，当 $x = 0$ 时，有：
$$
\int_{0}^{x} f(t) \mathrm{d} t + \int_{0}^{x} t f(x – t) \mathrm{d} t = ax^{2} \Rightarrow
$$
$$
\int_{0}^{0} f(t) \mathrm{d} t + \int_{0}^{0} t f(x – t) \mathrm{d} t = 0 \Rightarrow
$$
$$
0 = 0 \Rightarrow
$$
$$
f(0) = 0.
$$
即：
$$
f(0) = 2a + C = 0 \Rightarrow
$$
$$
C = -2a.
$$
于是：
$$
f(x) = 2a – 2a e^{-x} \Rightarrow
$$
$$
f(x) = 2a (1-e^{-x}).
$$
第 $(Ⅱ)$ 问
结合第 $(Ⅰ)$ 问，可得：
$$
\frac{\int _{0} ^{1} 2a (1-e^{-x}) \mathrm{d} x}{1 – 0} = 1
\Rightarrow
$$
$$
\int _{0} ^{1} 2a (1-e^{-x}) \mathrm{d} x = 1
\Rightarrow
$$
$$
2a \int _{0} ^{1} \mathrm{d} x – 2a \int_{0}^{1} e^{-x} \mathrm{d} x = 1
\Rightarrow
$$
$$
2a – 2a \cdot(- e^{-x} |_{0}^{1}) = 1 \Rightarrow
$$
$$
2a – 2a(\frac{-1}{e} + 1) \Rightarrow
$$
$$
2a – 2a(1 – \frac{1}{e}) \Rightarrow
$$
$$
2a – 2a + \frac{2a}{e} = 1 \Rightarrow
$$
$$
\frac{2a}{e} = 1 \Rightarrow
$$
$$
2a = e \Rightarrow
$$
$$
a = \frac{e}{2}.
$$