下册 8.1 二重积分 第44题

\section*{解题过程：} （1）方法 1：由分部积分法得 $$ \begin{aligned} \int_{0}^{x}\left(\int_{0}^{t} f(u) \mathrm{d} u\right) \mathrm{d} t & =\left.\left(t \int_{0}^{t} f(u) \mathrm{d} u\right)\right|_{0} ^{x}-\int_{0}^{x} t f(t) \mathrm{d} t=x \int_{0}^{x} f(u) \mathrm{d} u-\int_{0}^{x} t f(t) \mathrm{d} t \\ & =\int_{0}^{x} x f(u) \mathrm{d} u-\int_{0}^{x} t f(t) \mathrm{d} t=\int_{0}^{x} x f(t) \mathrm{d} t-\int_{0}^{x} t f(t) \mathrm{d} t=\int_{0}^{x}(x-t) f(t) \mathrm{d} t \end{aligned} $$ 方法 2：交换积分顺序得 所以 $$ \begin{aligned} \int_{0}^{x}\left(\int_{0}^{t} f(u) \mathrm{d} u\right) \mathrm{d} t= & \int_{0}^{x}\left(\int_{u}^{x} f(u) \mathrm{d} t\right) \mathrm{d} u=\int_{0}^{x}(x-u) f(u) \mathrm{d} u=\int_{0}^{x}(x-t) f(t) \mathrm{d} t \\ & \int_{0}^{x}\left(\int_{0}^{t} f(u) \mathrm{d} u\right) \mathrm{d} t=\int_{0}^{x}(x-t) f(t) \mathrm{d} t \end{aligned} $$ 方法3：令 $F(x)=\int_{0}^{x} f(u)(x-u) \mathrm{d} u-\int_{0}^{x}\left(f_{0}^{u} f(x) \mathrm{d} x\right) \mathrm{d} u$ ，则 $F(0)=0, F^{\prime}(x) \equiv 0$ 。从而 $F(x) \equiv 0$ ，即 $$ \int_{0}^{x} f(t)(x-t) \mathrm{d} t=\int_{0}^{x}\left(\int_{0}^{t} f(u) \mathrm{d} u\right) \mathrm{d} t $$ （2）交换积分次序得 $\int_{a}^{b} \mathrm{~d} x \int_{a}^{x} f(y) \mathrm{d} y=\int_{a}^{b} f(y) \mathrm{d} y \int_{y}^{b} \mathrm{~d} x=\int_{a}^{b} f(y)(b-y) \mathrm{d} y=\int_{a}^{b} f(x)(b-x) \mathrm{d} x$ ． （3）记 $F(x, y)=(y-x)^{n} f(x), D=\{(x, y) \mid a \leqslant x \leqslant y, a \leqslant y \leqslant b\}$ ． 由于 $f(t)$ 为连续函数，故 $F(x, y)$ 在 $D$ 上连续，从而在 $D$ 上可积。而对每个 $y \in[a, b], \int_{a}^{y} F(x, y) \mathrm{d} x$ 存在，从而累次积分 $\int_{a}^{b} \mathrm{~d} y \int_{a}^{y} F(x, y) \mathrm{d} x$ 也存在。同理 $\int_{a}^{b} \mathrm{~d} x \int_{a}^{x} F(x, y) \mathrm{d} y$ 也存在。于是 $$ \begin{gathered} \iint_{D} F(x, y) \mathrm{d} x \mathrm{~d} y=\int_{a}^{b} \mathrm{~d} y \int_{a}^{y} F(x, y) \mathrm{d} x=\int_{a}^{b} \mathrm{~d} x \int_{x}^{b} F(x, y) \mathrm{d} y . \\ \int_{a}^{b} \mathrm{~d} x \int_{x}^{b} F(x, y) \mathrm{d} y=\int_{a}^{b} f(x) \mathrm{d} x \int_{x}^{b}(y-x)^{n} \mathrm{~d} y=\int_{a}^{b} f(x)\left(\left.\frac{1}{n+1}(y-x)^{n+1}\right|_{x} ^{b}\right) \mathrm{d} x=\frac{1}{n+1} \int_{a}^{b} f(x)(b-x)^{n+1} \mathrm{~d} x . \\ \text { 即 } \quad \int_{a}^{b} \mathrm{~d} y \int_{a}^{y}(y-x)^{n} f(x) \mathrm{d} x=\frac{1}{n+1} \int_{a}^{b}(b-x)^{n+1} f(x) \mathrm{d} x . \end{gathered} $$ （4）设 $F(x)=\int_{0}^{x} f(t) \mathrm{d} t$ ，故 $\int_{x}^{y} f(t) \mathrm{d} t=F(y)-F(x)$ ，且有 $$ \int_{0}^{1} \mathrm{~d} x \int_{x}^{1} \mathrm{~d} y \int_{x}^{y} f(x) f(y) f(z) \mathrm{d} z=\int_{0}^{1} f(x) \mathrm{d} x \int_{x}^{1} f(y)(F(y)-F(x)) \mathrm{d} y . $$ 交换积分顺序得 $$ \begin{aligned} & \int_{0}^{1} f(x) \mathrm{d} x \int_{x}^{1} f(y)(F(y)-F(x)) \mathrm{d} y \\ & =\int_{0}^{1} f(y) \mathrm{d} y \int_{0}^{x} f(x)(F(y)-F(x)) \mathrm{d} x=\int_{0}^{1} f(y) \mathrm{d} y \int_{0}^{x}(F(y)-F(x)) \mathrm{d}(F(x)) \\ & =\int_{0}^{1} f(x)\left(F(y) F(x)-\left.\frac{1}{2} F(x) F(y)\right|_{0} ^{x}\right) \mathrm{d} x=\frac{1}{2} \int_{0}^{1} f(x) F^{2}(x) \mathrm{d} x \\ & =\frac{1}{2} \int_{0}^{1} F^{2}(x) \mathrm{d} F(x)=\left.\frac{1}{6} F^{3}(x)\right|_{0} ^{1}=\frac{1}{6}\left(\int_{0}^{1} f(t)\right)^{3} \end{aligned} $$