题号:1875    题型:单选题    来源:2012年全国硕士研究生招生考试试题
类型:考研真题
设 $I_{k}=\int_{0}^{k \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x(k=1,2,3)$, 则有 ( )
$A.$ $I_{1} < I_{2} < I_{3}$. $B.$ $I_{3} < I_{2} < I_{1}$. $C.$ $I_{2} < I_{3} < I_{1}$. $D.$ $I_{2} < I_{1} < I_{3}$.
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答案:
D

解析:

解 $\mathrm{I}_{2}=\int_{0}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x=\int_{0}^{\pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x+\int_{\pi}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x=\mathrm{I}_{1}+\int_{-\pi}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x$ 又 $\pi < x < 2 \pi$ 时 $\mathrm{e}^{x^{2}} \sin x < 0$
故 $\int_{\pi}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x < 0 \quad$ 故 $\mathrm{I}_{2} < \mathrm{I}_{1}$
$\mathrm{I}_{3}=\int_{0}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x=\int_{0}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x+\int_{2 \pi}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x=\mathrm{I}_{2}+\int_{2 \pi}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x$
又 $2 \pi < x < 3 \pi$ 时 $\mathrm{e}^{x^{2}} \sin x > 0$
故 $\int_{2 \pi}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x > 0$ 故 $\mathrm{I}_{2} < \mathrm{I}_{3}$.
$I_{3}=\int_{0}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x=\int_{0}^{\pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x+\int_{\pi}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x=\mathrm{I}_{1}+\int_{\pi}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x$
$\int_{\pi}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x=\int_{\pi}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x+\int_{2 \pi}^{3 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x$
$=\int_{\pi}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x+\int_{\pi}^{2 \pi} \mathrm{e}^{(t+\pi)^{2}} \sin (t+\pi) \mathrm{d}(t+\pi)$
$=\int_{\pi}^{2 \pi} \mathrm{e}^{x^{2}} \sin x \mathrm{~d} x-\int_{\pi}^{2 \pi} \mathrm{e}^{(x+\pi)^{2}} \sin x \mathrm{~d} x=\int_{\pi}^{2 \pi}\left[\mathrm{e}^{x^{2}}-\mathrm{e}^{(x+\pi)^{2}}\right] \sin x \mathrm{~d} x > 0$
$\therefore \mathrm{I}_{3} > \mathrm{I}_{1}$
综上 $\mathrm{I}_{3} > \mathrm{I}_{1} > \mathrm{I}_{2}$. 故应选 D.

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