新讲 第13章 重积分 第1题
📝 题目
例 1 设一元函数 $f$ 在 $\left\lbrack {a,b}\right\rbrack$ 连续,一元函数 $g$ 在 $\left\lbrack {c,d}\right\rbrack$ 连续, 试证
(1) $\displaystyle{\iint }_{\left\lbrack {a,b}\right\rbrack \times \left\lbrack {c,d}\right\rbrack }f\left( x\right) g\left( y\right) \mathrm{d}\left( {x,y}\right) = {\int }_{a}^{b}f\left( x\right) \mathrm{d}x{\int }_{c}^{d}g\left( y\right) \mathrm{d}y$ ;
(2) $\displaystyle{\iint }_{\left\lbrack {a,b}\right\rbrack \times \left\lbrack {a,b}\right\rbrack }f\left( x\right) f\left( y\right) \mathrm{d}\left( {x,y}\right) = {\left( {\int }_{a}^{b}f\left( x\right) \mathrm{d}x\right) }^{2}$ .
💡 答案解析
证明 把重积分化为累次积分, 我们得到
(1) $\displaystyle{\iint }_{\left\lbrack {a,b}\right\rbrack \times \left\lbrack {c,d}\right\rbrack }f\left( x\right) g\left( y\right) \mathrm{d}\left( {x,y}\right)$
$$ = {\int }_{a}^{b}\left( {{\int }_{c}^{d}f\left( x\right) g\left( y\right) \mathrm{d}y}\right) \mathrm{d}x $$
$$ = {\int }_{a}^{b}\left( {f\left( x\right) {\int }_{c}^{d}g\left( y\right) \mathrm{d}y}\right) \mathrm{d}x $$
$$ = {\int }_{a}^{b}f\left( x\right) \mathrm{d}x{\int }_{c}^{d}g\left( y\right) \mathrm{d}y; $$
(2) $\mathop{\iint }\limits_{{\left\lbrack {a,b}\right\rbrack \times \left\lbrack {a,b}\right\rbrack }}f\left( x\right) f\left( y\right) \mathrm{d}\left( {x,y}\right)$
$$ = {\int }_{a}^{b}f\left( x\right) \mathrm{d}x{\int }_{a}^{b}f\left( y\right) \mathrm{d}y = {\left( {\int }_{a}^{b}f\left( x\right) \mathrm{d}x\right) }^{2}. $$