[裴礼文数学分析中的典型问题与方法习题参考解答]4.3.1

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证明:

(1). $\dps{\sqrt{2}e^{-\frac{1}{2}}<\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} e^{-x^2}\rd x<\sqrt{2}}$;

(2). $\dps{0<\frac{\pi}{2}-\int_0^\frac{\pi}{2} \frac{\sin x}{x}\rd x<\frac{\pi^3}{144}}$;

(3). $\dps{\frac{2}{9}\pi^2\leq \int_\frac{\pi}{6}^\frac{\pi}{2}\frac{2x}{\sin x}\rd x\leq \frac{4}{9}\pi^2}$.

证明:

(1). 由 $$\bex \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} e^{-x^2}\rd x =2\int_0^\frac{1}{\sqrt{2}} e^{-x^2}\rd x,\quad 0<x<\frac{1}{2}\ra e^{-\frac{1}{2}}<e^{-x^2}<1 \eex$$ 知 $$\bex \sqrt{2}e^{-\frac{1}{2}} =2e^{-\frac{1}{2}}\cdot \frac{1}{\sqrt{2}} <\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} e^{-x^2}\rd x<2\cdot \frac{1}{\sqrt{2}}=\sqrt{2}. \eex$$

(2). 由 $$\bex x-\frac{x^3}{3!}<\sin x<x,\quad 0<x<\frac{\pi}{2} \eex$$ 知 $$\bex 1-\frac{x^2}{6}<\frac{\sin x}{x}<1\ra \frac{\pi}{2}-\frac{\pi^3}{144}<\int_0^\frac{\pi}{2}\frac{\sin x}{x}\rd x<\frac{\pi}{2}. \eex$$

(3). 由 $$\bex \frac{\pi}{6}\leq x\leq \frac{\pi}{2}\ra \frac{1}{2}\leq \sin x\leq 1\ra 1\leq \frac{1}{\sin x}\leq 2 \ra 2x\leq \frac{2x}{\sin x}\leq 4x \eex$$ 知 $$\bex \frac{2\pi^2}{9}=x^2|_{\frac{\pi}{6}}^\frac{\pi}{2} \leq \int_\frac{\pi}{6}^\frac{\pi}{2} \frac{2x}{\sin x}\rd x\leq 2x^2|_{\frac{\pi}{6}}^\frac{\pi}{2} =\frac{4\pi^2}{9}. \eex$$