随笔分类 - 微积分(A)
摘要：\[ \begin{aligned} &\text{Part I.} \\ &\int_0^{A}\frac{f'(x)}{f^2(x)}dx=-\frac{1}{f(x)} \bigg|_0^A=\frac{1}{f(0)}-\frac{1}{f(A)} \\ &\int_{0}^{+\infty
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摘要：判断敛散性：\(\int_0^{+\infty} \frac{\sin^2x}{x^p}dx\) \[ \begin{aligned} &当 p \le 0 时显然发散 \\ &\int_0^{+\infty} \frac{\sin^2x}{x^p}dx= \int_0^{1} \frac{\sin
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摘要：计算：\(\int_0^{+\infty}\frac{\ln x}{(1+x)^2}dx\) \[ \begin{aligned} I&=\int_0^{+\infty} \frac{\ln x}{(1+x)^3}dx \\ &=-\int_0^{+\infty} \frac{\ln \frac{1
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摘要：\[ \begin{aligned} &F(x)=\int_0^{x}f(t)dt,F'(x)=f(x) \\ &F'(x)=f(x) \le \sqrt{1+2F(x)} \\ &\int_0^{x}\frac{F'(t)}{\sqrt{1+2F(t)}}dt \le \int_0^xdx=x \
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摘要：求：\(\int_0^1\frac{\ln(1+x)}{1+x^2}dx\) \[ I=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx \\ \xlongequal{x=\tan t} \int_0^{\frac{\pi}{4}} \frac{\ln(1+\tan t)}{\se
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摘要：微分方程习题课 小技巧：\(h+\sqrt{h^2-1}=g \Rightarrow (g-h)^2=\sqrt{h^2-1} \Rightarrow g^2-2gh+{\color{red}{h^2}}={\color{red}{h^2}}-1\) \[ \begin{aligned} &p=y'
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摘要：广义积分习题课————证明题 1 \[ \int_1^{+\infty}\sin x^2dx\sim \int_1^{+\infty} \sin x\frac{dx}{\sqrt{x}} 收敛(\frac{1}{2}<1),\lim_{x \to +\infty} \sin x^2:DNE \\ 非
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摘要：广义积分习题课————判断敛散性 1 \[ \lim_{x \to +\infty} \frac{\frac{x\ln x}{\sqrt{x^5+1}}}{\frac{1}{x^{\frac{5}{4}}}}=0,\int _1^{+\infty} \frac{dx}{x^{\frac{5}{4}}
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摘要：广义积分习题课————计算下列广义积分的值 1 \[ \int_0^{+\infty} \frac{dx}{(1+5x^2)\sqrt{1+x^2}} \xlongequal{x=\tan t} \int_0^{\frac{\pi}{2}} \frac{\sec t dt}{1+5\tan^2 t}
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摘要：计算：\(\int_{-1}^{1}\frac{x^{2020}+1}{2020^x+1}dx\) 考虑：\(\int_{-a}^{a} f(x)dx=\int_0^{a}f(x)dx+\int_{-a}^{0}f(x)dx=\int_0^a[f(x)+f(-x)]dx\) \[ \int_{-1}
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摘要：1 设切于 \((x_0,\ln x_0)\)，则 \(l:y=\frac{1}{x_0}(x-x_0)+\ln x_0(2 \le x_0 \le 6)\) \[ \begin{aligned} S=S(x_0)=&\int_2^6 \left(\frac{x-x_0}{x_0}+\ln x_0-
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摘要：若 \(f(x) \in C[0,\pi]\)，求证：\(\lim_{n \to \infty}\int_0^\pi f(x)|\sin nx|dx=\frac{2}{\pi}\int_0^\pi f(x)dx\) \[ \begin{aligned} \int_0^\pi f(x) |\sin n
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摘要：计算：\(\int_0^\pi \sin^6x\cos^2xdx\) 首先有： \[ \int \sin^6x\cos^2 x dx=\int (\sin^6x-\sin^8x)dx \\ \] 其次： \[ \begin{aligned} I_n =&\int \sin^ndx \\ =&-\in
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摘要：设 \(f(x) \in C[0,\pi]\)，且 \(\int_0^\pi f(x)dx=0,\int_0^\pi f(x)\cos xdx=0\) 求证：\(\exists \zeta_1,\zeta_2 \in (0,\pi),\zeta_1 \ne \zeta_2,s.t.f(\zeta_1
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摘要：比大小： \(I_1=\int_0^{\frac{\pi}{2}}\sin(\sin x)dx \\I_2=\int_0^{\frac{\pi}{2}}\cos(\sin x)dx\) \[ \begin{aligned} &\begin{cases} \sin x+\cos x \le \sqrt
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摘要：求：\(\int \sin^{n}dx\) \[ \begin{aligned} I_n =&\int \sin^ndx \\ =&-\int \sin^{n-1}d\cos x \\ =&-\sin^{n-1}x\cos x+\int \cos x(n-1)\sin^{n-2}\cos xdx \
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摘要：设 \(f(x) \in C^{1}[a,b]\)，且 \(f(a)=0\) 求证：\(\int_{a}^{b}f^2(x)dx \le \frac{(b-a)^2}{2}\int_a^b[f'(x)]^2dx\) \[ \begin{aligned} &\because f(x)=f(x)-f(a
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摘要：计算：\(\int_{-\pi}^{\pi} \frac{x\sin^3x}{1+e^x}dx\) \[ \int_{-\pi}^{\pi} \frac{x\sin^3x}{1+e^x}dx \\ =\int_{-\pi}^{0} \frac{x\sin^3x}{1+e^x}dx +\int_{0}
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摘要：计算：\(\int_{0}^{1}\frac{dx}{(x+1)\sqrt{x^2+1}}\) 试试散装复变 \[ \int_{0}^{1}\frac{dx}{(x+1)\sqrt{x^2+1}} \\ =\int_{0}^{\frac{\pi}{4}}\frac{\sec^2tdt}{(1+\ta
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摘要：求：\(\lim_{x \to 0^+}\frac{x \ln \sin x-\sin x \ln x}{x^3\ln x}\) \[ \lim_{x \to 0^+}\frac{x \ln \sin x-\sin x \ln x}{x^3\ln x} \\ =\lim_{x \to 0^+}\fr
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