方企勤 第三章 一元函数积分学 第5题
📝 题目
例 5 设 $f\left( x\right) = \left\{ \begin{array}{ll} 1 + {x}^{2}, & x < 0, \\ {\mathrm{e}}^{-x}, & x \geq 0, \end{array}\right.$ 求 $\displaystyle{\int }_{1}^{3}f\left( {x - 2}\right) \mathrm{d}x$ .
💡 答案解析
解法 1 作变量代换 $t = x - 2$ ,则
$$ {\int }_{1}^{3}f\left( {x - 2}\right) \mathrm{d}x = {\int }_{-1}^{1}f\left( t\right) \mathrm{d}t = {\int }_{-1}^{0}f\left( t\right) \mathrm{d}t + {\int }_{0}^{1}f\left( t\right) \mathrm{d}t $$
$$ = {\int }_{-1}^{0}\left\lbrack {1 + {t}^{2}}\right\rbrack \mathrm{d}t + {\int }_{0}^{1}{\mathrm{e}}^{-t}\mathrm{\;d}t = 1 + {\left. \frac{1}{3} - {\mathrm{e}}^{-t}\right| }_{0}^{1} $$
$$ = 1 + \frac{1}{3} - \frac{1}{\mathrm{e}} + 1 = \frac{7}{3} - \frac{1}{\mathrm{e}}. $$
解法 2 因为
$$ f\left( x\right) = \left\{ \begin{array}{ll} 1 + {x}^{2}, & x < 0 \\ {\mathrm{e}}^{-x}, & x \geq 0 \end{array}\right. $$
$$ \Rightarrow f\left( {x - 2}\right) = \left\{ \begin{array}{ll} 1 + {\left( x - 2\right) }^{2}, & x < 2, \\ {\mathrm{e}}^{2 - x}, & x \geq 2, \end{array}\right. $$
所以
$$ {\int }_{1}^{3}f\left( {x - 2}\right) \mathrm{d}x = {\int }_{1}^{2}\left\lbrack {1 + {\left( x - 2\right) }^{2}}\right\rbrack \mathrm{d}x + {\int }_{2}^{3}{\mathrm{e}}^{2 - x}\mathrm{\;d}x $$
$$ = {\left. 1 - \frac{1}{3}{\left( x - 2\right) }^{3}\right| }_{1}^{2} - {\left. {\mathrm{e}}^{2 - x}\right| }_{2}^{3} $$
$$ = 1 + \frac{1}{3} - \frac{1}{\mathrm{e}} + 1 = \frac{7}{3} - \frac{1}{\mathrm{e}}. $$
评注 以上两种解法, 解法 1 容易操作些, 先通过变量代换分别将被积函数的积分变量和积分上、下限进行替换和变限, 再将积分区间按分段函数的方式分段,并将积分写成分段相加,最后代入各段相应的表达式, 就便于积分了.