对变上限积分 $\int_{0}^{x}$ $t f(x – t)$ $\mathrm{d} t$ 进行求导运算

一、题目

对变上限积分：

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

进行求导运算的结果是什么？

二、解析

令 $\textcolor{cyan}{u}$ $\textcolor{cyan}{=}$ $\textcolor{cyan}{x}$ $\textcolor{cyan}{-}$ $\textcolor{cyan}{t}$, 则：

$$
t = x – u.
$$

$$
\textcolor{cyan}{
\mathrm{d} t = – \mathrm{d} u}.
$$

又由于 $t$ $\in$ $(0, x)$, 因此，$\textcolor{cyan}{u}$ $=$ $x$ $-$ $t$ $\textcolor{cyan}{\in}$ $\textcolor{cyan}{(x, 0)}$.

Next

于是：

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

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

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

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

$$
\textcolor{orange}{
x \int_{0}^{x} f(u) \mathrm{d} u – \int_{0}^{x} u \cdot f(u) \mathrm{d} u}.
$$

Next

从而：

$$
\textcolor{orange}{
\Big [\int_{0}^{x} t f(x – t) \mathrm{d} t \Big]^{\prime}} =
$$

$$
\Big[ x \int_{0}^{x} f(u) \mathrm{d} u – \int_{0}^{x} u \cdot f(u) \mathrm{d} u \Big]^{\prime} =
$$

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

$$
\textcolor{red}{
\int_{0}^{x} f(u) \mathrm{d} u}.
$$

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