108年一般警察三等考試-消防警察人員-微積分詳解

108年一般警察人員考試

等      別：三等考試
類科別：消防警察人員
科       目：微積分

解：
使用萊布尼茲積分法則(Leibniz Integral Rule)來求解
$$F(x)= \int_{\sin x}^{\cos x} e^{t^2+xt}dt \\ \Rightarrow F'(x)= \int_{\sin x}^{\cos x}  \frac{\partial}{\partial x} e^{t^2+xt}dt +e^{\cos^2x+x\cos x} \frac{d}{d x} \cos x- e^{\sin^2x+x\sin x}\frac{d}{d x} \sin x \\= \int_{\sin x}^{\cos x}  t e^{t^2+xt}dt - \sin xe^{\cos^2x+x\cos x}- \cos xe^{\sin^2x+x\sin x} \\ \Rightarrow F'(0)=\int_0^1 te^{t^2}dt-0-e^0 = \int_0^1 te^{t^2}dt-1 = \left. \left[ \frac{1}{2}e^{t^2}\right] \right|_0^1-1\\ =\frac{1}{2}(e^1-e^0)-1= \bbox[red, 2pt]{\frac{1}{2}e-\frac{3}{2}}$$

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解：

$$ \lim_{x \rightarrow 0} \left(  \frac{\tan 2x}{x^3} + \frac{a}{x^2} + \frac{\sin bx}{x} \right) =  \lim_{x \rightarrow 0}   \frac{ax+x^2\sin (bx)+\tan (2x)}{x^3}= \lim_{x\to 0} \frac{f(x)}{g(x)} \\ 由於\lim_{x\to 0}f(x) = \lim_{x\to 0}g(x) =0，\text{由羅必達法則(L'Hopital's Rule)}可知 \lim_{x\to 0} \frac{f(x)}{g(x)}=\lim_{x\to 0} \frac{f'(x)}{g'(x)}\\ =  \lim_{x \rightarrow 0}   \frac{a+2x\sin (bx)+ bx^2\cos(bx)+2\sec^2 (2x)}{3x^2} \Rightarrow  \begin{cases}f'(x) =a+2x\sin (bx)+ bx^2\cos(bx)+2\sec^2 (2x) \\g'(x) =3x^2\end{cases} \\  \Rightarrow  \begin{cases}f'(0)=a+2\\g'(0) =0\end{cases}  \Rightarrow \bbox[red, 2pt]{a=-2} \Rightarrow \lim_{x\to 0} \frac{f(x)}{g(x)}=\lim_{x\to 0} \frac{f'(x)}{g'(x)} =\lim_{x\to 0} \frac{f''(x)}{g''(x)} \\=\lim_{x\to 0} \frac{2\sin(bx) +4bx\cos(bx)-b^2x^2\sin(bx) +8\sec^2(2x) \tan(2x)}{6x}  \Rightarrow  \begin{cases}f''(0)= 0\\g ''(0)=0\end{cases} \\  \Rightarrow \lim_{x\to 0} \frac{f''(x)}{g''(x)} =\lim_{x\to 0} \frac{f'''(x)}{g'''(x)} \\=\lim_{x\to 0} \frac{6b\cos(bx)  -6b^2x\sin(bx)  -b^3x^2\cos(bx) +32\sec^2(2x)\tan^2(2x) +16\sec^4(2x)}{6} \\  \Rightarrow  \begin{cases}f'''(0)= 6b+16\\ g'''(x)=6\end{cases} ,由於\lim_{x\to 0} \frac{f(x)}{g(x)}=0 =\lim_{x\to 0} \frac{f'''(x)}{g'''(x)}=\frac{6b+16}{6}  \Rightarrow 6b+16=0 \Rightarrow \bbox[red, 2pt]{b=-\frac{8}{3}}$$

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解：$$\frac{x^2}{9} + \frac{y^2}{16} =1 \Rightarrow  \begin{cases} x = 3\cos \theta\\ y=4\sin \theta\end{cases} ,0\le \theta\le 2\pi \\ \Rightarrow P(3\cos \theta, 4\sin \theta)至L:x+y=6  的距離\quad d(P,L)= \left|  \frac{3\cos \theta+4\sin \theta-6}{ \sqrt{1+1} }  \right|\\ =\left|  \frac{5(\frac{3}{5}\cos \theta+\frac{4}{5}\sin \theta)-6}{ \sqrt{2} }  \right|= \left|  \frac{5\sin(\theta+\alpha)-6}{ \sqrt{2} }  \right|  \Rightarrow  \frac{1}{ \sqrt{2} }\le d(P,L)\le  \frac{11}{ \sqrt{2} } \\  \Rightarrow 最短距離為\bbox[red, 2pt]{\frac{1}{ \sqrt{2} }}$$

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解：
$$y= \frac{x^3}{12} + \frac{1}{x}  \Rightarrow y'=\frac{x^2}{4} - \frac{1}{x^2 } \Rightarrow (y')^2= \frac{x^4}{16} - \frac{1}{2}  + \frac{1}{x^4}\\ \Rightarrow 曲線長度= \int_1^4  \sqrt{1+(y')^2}  dx=\int_1^4  \sqrt{ \frac{x^4}{16} + \frac{1}{2}  + \frac{1}{x^4}}  dx= \int_1^4  \sqrt{ (\frac{x^2}{4} + \frac{1}{x^2})^2}dx\\ = \int_1^4 \left(\frac{x^2}{4} + \frac{1}{x^2}\right) dx =\left. \left[ \frac{x^3}{12}- \frac{1}{x}\right] \right|_1^4 =\left( \frac{16}{3} -\frac{1}{4} \right)- \left( \frac{1}{12} -1\right)= \bbox[red, 2pt]{\frac{18}{3}}$$

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解：
(一)$$令 \begin{cases}u = \cos^{n-1}x\\dv=\cos{x}dx\end{cases}  \Rightarrow 令 \begin{cases}du = (n-1)\cos^{n-2}x(-\sin{x})dx\\ v=\sin{x}\end{cases}\\ \Rightarrow \int \cos^n{x}\;dx= uv-\int vdu = \sin{x}\cos^{n-1}x+ (n-1)\int \sin^2{x}\cos^{n-2}{x}\;dx \\= \sin{x}\cos^{n-1}x+ (n-1)\int (1-\cos^2{x})\cos^{n-2}{x}\;dx \\ = \sin{x}\cos^{n-1}x+ (n-1)\int \cos^{n-2}{x}\;dx -(n-1)\int \cos^n{x}\;dx \\  \Rightarrow n\int \cos^n(x)\;dx = \sin{x}\cos^{n-1}x+ (n-1)\int \cos^{n-2}{x}\;dx\\  \Rightarrow \int \cos^n{x}\;dx = \frac{\sin{x}\cos^{n-1}x} {n} + \frac{n-1}{n} \int \cos^{n-2}{x}\;dx,\text{for }n\ge 2\\ 故得證。$$(二)$$\int_0^{\pi/2}{\cos^{10}x\;dx}= \left.\left[\frac{\cos^9{x}\sin{x}}{10}\right]\right|_0^{\pi/2} + \frac{9}{10} \int_0^{\pi/2} \cos^8{x}\;dx =\frac{9}{10} \int_0^{\pi/2} \cos^8{x}\;dx \\ =\frac{9}{10}\left( \left. \left[ \frac{\cos^7{x}\sin{x}}{8}\right]\right|_0^{\pi/2} +\frac{7}{8}\int_0^{\pi/2}\cos^6{x}\;dx \right) = \frac{9}{10}  \cdot \frac{7}{8}\int_0^{\pi/2}\cos^6{x}\;dx \\ =\frac{9}{10}  \cdot \frac{7}{8} \left( \left. \left[ \frac{\cos^5{x}\sin{x}}{6}\right]\right|_0^{\pi/2} +\frac{5}{6}\int_0^{\pi/2} \cos^4{x}\;dx \right) = \frac{9}{10}  \cdot \frac{7}{8}\cdot \frac{5}{6}\int_0^{\pi/2} \cos^4{x}\;dx  \\ = \frac{9}{10}  \cdot \frac{7}{8}\cdot \frac{5}{6} \left( \left. \left[ \frac{\cos^3{x}\sin{x}}{4}\right]\right|_0^{\pi/2} +\frac{3}{4}\int_0^{\pi/2} \cos^2{x}\;dx \right) = \frac{9}{10}  \cdot \frac{7}{8}\cdot \frac{5}{6}  \cdot \frac{3}{4}\int_0^{\pi/2} \cos^2{x}\;dx \\ = \frac{9}{10}  \cdot \frac{7}{8}\cdot \frac{5}{6}  \cdot \frac{3}{4}\left( \left. \left[ \frac{\cos{x}\sin{x}}{4}\right]\right|_0^{\pi/2} +\frac{1}{2}\int_0^{\pi/2} \cos^0{x}\;dx \right) \\=  \frac{9}{10}  \cdot \frac{7}{8}\cdot \frac{5}{6}  \cdot \frac{3}{4}\cdot \frac{1}{2}\int_0^{\pi/2} 1\;dx =\frac{9}{10}  \cdot \frac{7}{8}\cdot \frac{5}{6}  \cdot \frac{3}{4}\cdot \frac{1}{2} \cdot \frac{\pi}{2} = \bbox[red, 2pt]{\frac{63}{512} \pi}$$

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解：

$$\begin{cases}y=x^2+1\\ y=2x+1 \end{cases}  \Rightarrow 交點 \begin{cases}P(0,1)\\ Q(2,5)\end{cases} ，圖形見上；\\D=\pi\int_0^2(2x+1)^2-(x^2+1)^2\;dx =\pi\int_0^2-x^4+2x^2+4x\;dx =\pi \left. \left[ -\frac{1}{5}x^5+ \frac{2}{3}x^3 +2x^2 \right] \right|_0^2 \\ =\pi \left( -\frac{32}{5} + \frac{16}{3} +8 \right) = \bbox[red, 2pt]{\frac{104}{15}\pi}$$

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解：
$$y= \sqrt{4-x^2}  \Rightarrow y'= \frac{-x}{ \sqrt{4-x^2} }  \Rightarrow (y')^2= \frac{x^2}{4-x^2} \\ 旋轉曲面面積=2\pi \int_{-1}^1 y\sqrt{1+(y')^2}dx =2\pi \int_{-1}^1 \sqrt{4-x^2}\sqrt{1+\frac{x^2}{4-x^2}}dx  \\ =2\pi \int_{-1}^1 \sqrt{4-x^2}\sqrt{\frac{4}{4-x^2}}dx  =2\pi \int_{-1}^12\;dx = \bbox[red, 2pt]{8\pi}$$

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