115年政大應數轉學考-微積積分二詳解

 國立政治大學 115 學年度學士班轉學生招生考試試

考試科目 微積分 (二): 以積分為主
系所別 應用數學系二年級

解答:$$\textbf{(a) }u=2+\sqrt x \Rightarrow du={1\over 2\sqrt x} dx \Rightarrow \int_1^4 {\sqrt{2+\sqrt x} \over \sqrt x}\,dx=\int_3^4 2\sqrt u\,du = 2 \left. \left[ {2\over 3}u^{3/2} \right] \right|_3^4 \\ \qquad = {4\over 3} (8-3\sqrt 3) = \bbox[red, 2pt]{{32\over 3}-4\sqrt 3} \\\textbf{(b) }\cases{u =\ln(1+x) \\ dv=xdx} \Rightarrow \cases{du=dx/(1+x)  \\ v={1\over 2}x^2} \Rightarrow  \int_0^1 x\ln(1+x)\,dx \\\qquad= \left. \left[ {1\over 2}x^2 \ln(1+x) \right] \right|_0^1 - \int_0^1 {x^2\over 2(1+x)}\,dx  = {1\over 2}\ln 2-{1\over 2} \int_0^1 \left( x-1+{1\over x+1} \right)\,dx \\\qquad ={1\over 2}\ln 2-{1\over 2} \left. \left[ {1\over 2}x^2-x+ \ln(x+1) \right] \right|_0^1  ={1\over 2}\ln 2-{1\over 2} \left( -{1\over 2}+ \ln 2 \right)= \bbox[red, 2pt]{1\over 4} \\ \textbf{(c) } \int_0^1 {x\over x^2+3x+2} \,dx = \int_0^1 \left( {-1\over x+1}+{2\over x+2} \right)\,dx = \left. \left[  -\ln|x+1|+2\ln|x+2 \right] \right|_0^1 = \bbox[red, 2pt]{\ln{9\over 8}} \\\textbf{(d) } x=2\sin \theta \Rightarrow dx  =2\cos \theta \,d\theta \Rightarrow \int_0^1 {x^2\over \sqrt{4-x^2}}\,dx = \int_0^{\pi/6} {4\sin^2\theta\over 2\cos \theta}\cdot 2\cos \theta \,d\theta \\\qquad= \int_0^{\pi/6} 4\sin^2\theta =2\int_0^{\pi/6} (1-\cos 2\theta)\,d\theta =2 \left. \left[ \theta-{1\over 2} \sin 2\theta\right] \right|_0^{\pi/6}  = \bbox[red, 2pt]{{\pi \over 3}-{\sqrt 3\over 2}}$$

解答:$$I=\int_0^1 f(x)\,dx = \sum_{i=1}^n \int_{(i-1)/n}^{i/n} f(x)\,dx \\ 取\cases{u=f(x)\\ dv= dx} \Rightarrow \cases{du=f'(x) \\ 取v=x-(i-1)/n} \\\Rightarrow \int_{(i-1)/n}^{i/n} f(x)\,dx =  \left. \left[ \left( x-{i-1\over n} \right) f(x)\right] \right|_{(i-1)/n}^{i/n}-\int_{(i-1)/n}^{i/n} \left( x-{i-1\over n}  \right) f'(x)\,dx \\={1\over n} f \left( {i\over n} \right)-\int_{(i-1)/n}^{i/n} \left( x-{i-1\over n}  \right) f'(x)\,dx \\ \Rightarrow {1\over n} f \left( {i\over n} \right)-\int_{(i-1)/n}^{i/n} f(x)\,dx =\int_{(i-1)/n}^{i/n} \left( x-{i-1\over n}  \right) f'(x)\,dx \\ \Rightarrow {1\over n} \sum_{i=1}^n f \left( {i\over n} \right)-\int_0^1f(x)\,dx = \sum_{i=1}^n \int_{(i-1)/n}^{i/n} \left( x-{i-1\over n}  \right) f'(x)\,dx\\ 取x={i-1\over n}+{t\over n } \Rightarrow dx={1\over n}\,dt\\\qquad  \Rightarrow \int_{(i-1)/n}^{i/n} \left( x-{i-1\over n}  \right) f'(x)\,dx= \int_0^1 {t\over n}\cdot  f' \left( {i-1+t\over n} \right){1\over n}dt = {1\over n^2} \int_0^1 tf' \left( {i-1+t\over n} \right)\,dt \\ \Rightarrow  \sum_{i=1}^n \frac{1}{n^2} \int_0^1 t f'\left( \frac{i-1+t}{n} \right) dt = \frac{1}{n} \int_0^1 t \left[ \frac{1}{n} \sum_{i=1}^n f'\left( \frac{i-1+t}{n} \right) \right] dt \\ \Rightarrow  n \left[ \frac{1}{n} \sum_{i=1}^n f\left(\frac{i}{n}\right) - \int_0^1 f(x) dx \right] = \int_0^1 t \left[ \frac{1}{n} \sum_{i=1}^n f'\left( \frac{i-1+t}{n} \right) \right] dt \\ \Rightarrow \lim_{n\to \infty}   n \left[ \frac{1}{n} \sum_{i=1}^n f\left(\frac{i}{n}\right) - \int_0^1 f(x) dx \right] = \lim_{n \to \infty} \int_0^1 t \left[ \frac{1}{n} \sum_{i=1}^n f'\left( \frac{i-1+t}{n} \right) \right] dt\\ = \int_0^1 t \left( \int_0^1 f'(x) dx \right) dt =\int_0^1 t(f(1)-f(0))\,dt  =(f(1)-f(0)) \left. \left[ {t^2\over 2} \right] \right|_0^1 ={f(1)-f(0) \over 2}\; \bbox[red, 2pt]{故得證} \\取f(x)=x^9 \Rightarrow \int_0^1 f(x)\,dx ={1\over 10} \\\Rightarrow \cases{f(1)=1\\ f(0)=0} \Rightarrow \lim_{n \to \infty} n \left[ \frac{1}{n} \left\{ \left(\frac{1}{n}\right)^9 + \left(\frac{2}{n}\right)^9 + \cdots + \left(\frac{n}{n}\right)^9 \right\} - \frac{1}{10} \right] = \bbox[red, 2pt]{1\over 2}$$

解答:$$u=x-2  \Rightarrow I=\int_0^4 xe^{(x-2)^4} \,dx = \int_{-2}^2 (u+2)e^{u^4}\,du =\int_{-2}^2 ue^{u^4}\,du +\int_{-2}^22e^{u^4} \,du\\ ue^{u^4}為奇函數\Rightarrow \int_{-2}^2 ue^{u^4}\,du=0 \Rightarrow I=0  +\int_{-2}^22e^{u^4} \,du= 2\int_0^4  e^{(x-2)^4} \,dx =\bbox[red, 2pt]{2k}$$

解答:$${d^2\over dx^2 } \int_0^x \left( \int_1^{e^x} \sqrt{1+u^2} \,du\right)\,dt = {d\over dx} \int_1^{e^x} \sqrt{1+u^2}\,du = \bbox[red, 2pt]{e^x \sqrt{1+e^{2x}}}$$

解答:$$\int_0^\infty e^{ax} \cos x\,dx = \lim_{t\to \infty} \int_0^te^{ax} \cos x\,dx = \lim_{t\to \infty} \left. \left[ {e^{ax} \over a^2+1}(a\cos x+\sin x) \right] \right|_0^t \\=  \lim_{t \to \infty} \left( \frac{e^{at}}{a^2 + 1} (a \cos t + \sin t) \right) - \frac{a}{a^2 + 1} \\ \cases{a\gt 0 \Rightarrow e^{at} \to \infty \Rightarrow \text{diverge} \\ a=0 \Rightarrow e^{at}=1 \Rightarrow \text{diverge} \\ a\lt 0 \Rightarrow e^{at} \to 0\Rightarrow \text{ converge}} \Rightarrow \text{the integral converges iff } \bbox[red, 2pt]{a\lt 0} \\ a\lt 0 \Rightarrow \lim_{t \to \infty} \left( \frac{e^{at}}{a^2 + 1} (a \cos t + \sin t) \right) - \frac{a}{a^2 + 1}= \bbox[red, 2pt]{-{a\over a^2+1}}$$

解答:$$x=x^2 \Rightarrow x(x-1)=0 \Rightarrow \text{The intersection points are at $x = 0$ and $x = 1$}.  \\ \text{volume }V = 2\pi \int_0^1 \text{(radius)(height)} \,dx = 2\pi \int_0^1(2-x)(x-x^2)\,dx\\= 2\pi \int_0^1 (x^3-3x^2+2x)\, dx =\bbox[red, 2pt] {\pi\over 2}$$

解答:$$I=\int_0^1 \int_{\sin^{-1}y}^{\pi/2} \cos x\sqrt{1+\cos^2 x} \,dx \,dy =  \int_0^{\pi/2} \int_0^{\sin x} \cos x \sqrt{1 + \cos^2 x} \, dy \, dx  \\  =  \int_0^{\pi/2} \left. \left[ y \cos x \sqrt{1+\cos^2 x} \right] \right|_0^{\sin x} \,dx =  \int_0^{\pi/2} \sin x \cos x \sqrt{1 + \cos^2 x} \, dx  \\ u=1+\cos^2 x \Rightarrow du=-2\sin x\cos x\,dx \Rightarrow I= \int_2^1 \sqrt u \left( -{1\over 2} \right)\,du =  \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_1^2  \\= \bbox[red, 2pt]{2\sqrt 2-1\over 3}$$

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