## 张祖锦2023年数学专业真题分类70天之第15天
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323、 4、 (15 分) 计算不定积分. (1)、 (8 分) $\displaystyle \int \mathrm\{e\}^\{\sin x\}\sin 2x\mathrm\{ d\} x$; (新疆大学2023年数学分析考研试题) [不定积分 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} \int \mathrm\{e\}^\{\sin x\}\sin 2x\mathrm\{ d\} x =&2\int \sin x\cdot \mathrm\{e\}^\{\sin x\}\cos x\mathrm\{ d\} x =2\int \sin x\mathrm\{ d\} \mathrm\{e\}^\{\sin x\}\\\\ =&2\left\[\mathrm\{e\}^\{\sin x\}\sin x-\int \cos x\mathrm\{e\}^\{\sin x\}\mathrm\{ d\} x\right\]\\\\ =&2\mathrm\{e\}^\{\sin x\}(\sin x-1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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324、 (2)、 (7 分) $\displaystyle \int\frac\{x-2\}\{x^2-7x+12\}\mathrm\{ d\} x$. (新疆大学2023年数学分析考研试题) [不定积分 ]
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\begin\{aligned\} \mbox\{原式\}=&\int\left(\frac\{2\}\{x-4\}-\frac\{1\}\{x-3\}\right)\mathrm\{ d\} x =2\ln |x-4|-\ln |x-3|+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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325、 (2)、 (10 分) 计算不定积分 $\displaystyle I\_n=\int \frac\{x^\{2n-1\}\}\{x^n+1\}\mathrm\{ d\} x$, $\displaystyle n\in\mathbb\{N\}\_+$. (长安大学2023年数学分析考研试题) [不定积分 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} I\_n=&\int \frac\{x^n\}\{1+x^n\}\cdot x^\{n-1\}\mathrm\{ d\} x \stackrel\{x^n=t\}\{=\}\frac\{1\}\{n\}\int \frac\{t\}\{1+t\}\mathrm\{ d\} t\\\\ =&\frac\{1\}\{n\}\int \left(1-\frac\{1\}\{1+t\}\right)\mathrm\{ d\} t =\frac\{t-\ln (1+t)\}\{n\}+C =\frac\{x^n-\ln(1+x^n)\}\{n\}+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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326、 2、 (10 分) 求不定积分 $\displaystyle \int \frac\{x\ln x\}\{(1+x^2)^2\}\mathrm\{ d\} x$. (郑州大学2023年数学分析考研试题) [不定积分 ]
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https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
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\begin\{aligned\} \mbox\{原式\}=&\frac\{-1\}\{2\}\int \ln x\mathrm\{ d\} \frac\{1\}\{1+x^2\} \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\frac\{1\}\{2\}\left\[\frac\{\ln x\}\{1+x^2\}-\int \frac\{1\}\{x\}\cdot \frac\{1\}\{1+x^2\}\mathrm\{ d\} x\right\]\\\\ &=-\frac\{\ln x\}\{2(1+x^2)\}+\frac\{1\}\{2\}\int\left(\frac\{1\}\{x\}-\frac\{x\}\{1+x^2\}\right)\mathrm\{ d\} x\\\\ &=-\frac\{\ln x\}\{2(1+x^2)\}+\frac\{1\}\{2\}\ln x-\frac\{1\}\{4\}\ln(1+x^2)+C\\\\ &=\frac\{x^2\ln x\}\{2(1+x^2)\}-\frac\{1\}\{4\}\ln (1+x^2)+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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327、 1、 (15 分) 计算积分. (1)、 (7 分) $\displaystyle \int\frac\{1+x^4\}\{1+x^6\}\mathrm\{ d\} x$. (中山大学2023年数学分析考研试题) [不定积分 ]
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https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\frac\{1\}\{3\}\int \left\[\frac\{2\}\{1+x^2\}+\frac\{1+x^2\}\{1-x^2+x^4\}\right\]\mathrm\{ d\} x\\\\ =&\frac\{2\}\{3\}\arctan x+\frac\{1\}\{3\}\int \frac\{\frac\{1\}\{x^2\}+1\}\{\frac\{1\}\{x^2\}-1+x^2\}\mathrm\{ d\} x\\\\ =&\frac\{2\}\{3\}\arctan x+\frac\{1\}\{3\}\int \frac\{\mathrm\{ d\} \left(x-\frac\{1\}\{x\}\right)\}\{1+\left(x-\frac\{1\}\{x\}\right)^2\}\\\\ =&\frac\{2\}\{3\}\arctan x+\frac\{1\}\{3\}\arctan \left(x-\frac\{1\}\{x\}\right)+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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328、 (3)、 计算不定积分 $\displaystyle I=\int\frac\{\ln \sin x\}\{\sin^2x\}\mathrm\{ d\} x$. (重庆大学2023年数学分析考研试题) [不定积分 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} \mbox\{原式\}=&\int \ln \sin x\cdot \csc^2x\mathrm\{ d\} x =-\int \ln \sin x\mathrm\{ d\} \cot x\\\\ =&-\left\[\cot x\ln \sin x-\int\frac\{\cos x\}\{\sin x\}\cdot\cot x\mathrm\{ d\} x\right\]\\\\ =&-\cot x\ln \sin x+\int\frac\{1-\sin^2x\}\{\sin^2x\}\mathrm\{ d\} x\\\\ =&-\cot x\ln \sin x-\cot x-x+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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329、 (2)、 $\displaystyle \int \frac\{1\}\{\sqrt\{2+4x\}\}\mathrm\{ d\} x$. (重庆师范大学2023年数学分析考研试题) [不定积分 ]
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\begin\{aligned\} \mbox\{原式\}\stackrel\{\sqrt\{2+4x\}=t, x=\frac\{t^2-2\}\{4\}\}\{=\}\int \frac\{1\}\{t\}\cdot \frac\{t\}\{2\}\mathrm\{ d\} t=\frac\{t\}\{2\}+C =\frac\{\sqrt\{2+4x\}\}\{2\}+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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330、 2、 试问闭区间 $\displaystyle [a,b]$ 上的连续函数类 $\displaystyle C[a,b]$, 可导函数类 $\displaystyle D[a,b]$, 可积函数类 $\displaystyle R[a,b]$ 之间有什么关系? 请写出证明过程或举出反例. (北京工业大学2023年数学分析考研试题) [定积分 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle D(a,b)\subsetneq C[a,b]\subsetneq R[a,b]$. (1)、 $\displaystyle D(a,b)\subsetneq C[a,b]$. 设 $\displaystyle f\in D[a,b]$, 则 $\displaystyle \forall\ x\_0\in [a,b]$,
\begin\{aligned\} \lim\_\{h\to 0\}\frac\{f(x\_0+h)-f(x\_0)\}\{h\}=0\Rightarrow \lim\_\{h\to 0\}[f(x\_0+h)-f(x\_0)]=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle f\in C[a,b]$. 但由 $\displaystyle f(x)=|x|\in C[-1,1]$ 不可导知 $\displaystyle D(a,b)\subsetneq C[a,b]$. (2)、 $\displaystyle C[a,b]\subsetneq R[a,b]$. 设 $\displaystyle f\in C[a,b]$, 则 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上一致连续,
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta > 0,\mathrm\{ s.t.\}\forall\ x,x'\in [a,b], |x-x'| < \delta, |f(x)-f(x')| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle [a,b]$ 的分划 $\displaystyle T$, 使得 $\displaystyle \left\Vert T\right\Vert < \delta$, 则 $\displaystyle f$ 在 $\displaystyle T$ 的每个小区间的振幅 $\displaystyle \omega^f\_i < \varepsilon$. 于是
\begin\{aligned\} \sum\_i \omega^f\_i\Delta x\_i < \varepsilon \sum\_i \Delta x\_i=(b-a)\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\in R[a,b]$. 但由 $\displaystyle f(x)=[x]\in R[0,1]$ 不连续知 $\displaystyle C[a,b]\subsetneq R[a,b]$.跟锦数学微信公众号. [在线资料](
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331、 6、 证明函数 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上可积的充分必要条件是: 任给 $\displaystyle c\in (a,b)$, $\displaystyle f(x)$ 在 $\displaystyle [a,c]$ 和 $\displaystyle [c,b]$ 上都可积, 并证明
\begin\{aligned\} \int\_a^b f(x)\mathrm\{ d\} x=\int\_a^c f(x)\mathrm\{ d\} x+\int\_c^b f(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(北京工业大学2023年数学分析考研试题) [定积分 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Rightarrow$: 设 $\displaystyle f\in R[a,b]$, 则对 $\displaystyle \forall\ \varepsilon > 0$, 存在 $\displaystyle [a,b]$ 的分划 $\displaystyle T$, 使得 $\displaystyle \sum\_T \omega\_i\Delta x\_i < \varepsilon$. 在 $\displaystyle T$ 上再增加一个分点 $\displaystyle c$, 得到一个新的分划 $\displaystyle T^\star$. 由’区间越小蕴含振幅越小‘知
\begin\{aligned\} \sum\_\{T^\star\}\omega\_i^\star\Delta x\_i^\star\leq \sum\_T \omega\_i\Delta x\_i < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
记 $\displaystyle T'=T^\star\cap [a,c], T''=T^\star\cap [c,b]$, 则
\begin\{aligned\} \sum\_\{T'\}\omega\_i'\Delta x\_i'\leq \sum\_\{T^\star\}\omega\_i^\star < \varepsilon, \sum\_\{T''\}\omega\_i''\Delta x\_i''\leq \sum\_\{T^\star\}\omega\_i^\star < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\in R[a,c], f\in R[c,b]$, 且任取 $\displaystyle \xi\_i^\star\in [x\_\{i-1\}^\star,x\_i^\star]$ 后,
\begin\{aligned\} \sum\_\{T^\star\}f(\xi\_i^\star)\Delta x\_i^\star =\sum\_\{T'\}f(\xi\_i')\Delta x\_i' +\sum\_\{T''\}f(\xi\_i'')\Delta x\_i''. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle \left\Vert T^\star\right\Vert \to 0$ 即得 $\displaystyle \int\_a^b f(x)\mathrm\{ d\} x=\int\_a^c f(x)\mathrm\{ d\} x+\int\_c^b f(x)\mathrm\{ d\} x$. (2)、 $\displaystyle \Leftarrow$: 由 $\displaystyle f\in R[a,c], f\in R[c,b]$ 知对 $\displaystyle \forall\ \varepsilon > 0$, 存在 $\displaystyle [a,c]$ 的分划 $\displaystyle T'$, $\displaystyle [c,b]$ 的分划 $\displaystyle T''$, 使得
\begin\{aligned\} \sum\_\{T'\}\omega'\_i\Delta x'\_i < \frac\{\varepsilon\}\{2\}, \sum\_\{T''\}\omega''\_i\Delta x\_i'' < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle T=T'\cup T''$, 则它是 $\displaystyle [a,b]$ 的一个分划, 且
\begin\{aligned\} \sum\_T \omega\_i\Delta x\_i=\sum\_\{T'\}\omega'\_i\Delta x'\_i+\sum\_\{T''\}\omega''\_i\Delta x\_i'' < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\in R[a,b]$.跟锦数学微信公众号. [在线资料](
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332、 7、 已知 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, 记 $\displaystyle F(x)=\int\_a^x f(t)\mathrm\{ d\} t$. 证明: $\displaystyle F'(x)=f(x)$. (北京工业大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续知其一致连续:
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ s,t\in [a,b], |s-t| < \delta, |f(s)-f(t)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} 0 < |y-x| < \delta\Rightarrow \left|\frac\{F(y)-F(x)\}\{y-x\}-f(x)\right| =\left|\frac\{\int\_x^y [f(t)-f(x)]\mathrm\{ d\} t\}\{y-x\}\right| \leq \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle F'(x)=f(x)$.跟锦数学微信公众号. [在线资料](
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333、 2、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上可微, 且对于任何 $\displaystyle x\in (0,1)$, 由 $\displaystyle |f'(x)|\leq M$. 证明: 对于任何正整数 $\displaystyle n$, 有
\begin\{aligned\} \left|\int\_0^1 f(x)\mathrm\{ d\} x-\frac\{1\}\{n\}\sum\_\{i=1\}^n f\left(\frac\{i\}\{n\}\right)\right|\leq\frac\{M\}\{n\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle M$ 是一个与 $\displaystyle x$ 无关的常数. [张祖锦注: $\displaystyle \frac\{M\}\{n\}$ 可改进为 $\displaystyle \frac\{M\}\{2n\}$, 具体见参考解答] (北京科技大学2023年数学分析考研试题) [定积分 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\left|\int\_0^1 f(x)\mathrm\{ d\} x-\frac\{1\}\{n\}\sum\_\{i=1\}^n f\left(\frac\{i\}\{n\}\right)\right| =\left|\sum\_\{i=1\}^n \int\_\frac\{i-1\}\{n\}^\frac\{i\}\{n\} \left\[f(x)-f\left(\frac\{i\}\{n\}\right)\right\]\mathrm\{ d\} x\right|\\\\ \leq& \sum\_\{i=1\}^n\int\_\frac\{i-1\}\{n\}^\frac\{i\}\{n\} \left|f(x)-f\left(\frac\{i\}\{n\}\right)\right|\mathrm\{ d\} x =\sum\_\{i=1\}^n\int\_\frac\{i-1\}\{n\}^\frac\{i\}\{n\} |f'(\xi\_n)|\cdot \left|x-\frac\{i\}\{n\}\right|\mathrm\{ d\} x\\\\ \leq& M\sum\_\{i=1\}^n\int\_\frac\{i-1\}\{n\}^\frac\{i\}\{n\}\left|x-\frac\{i\}\{n\}\right|\mathrm\{ d\} x=\frac\{M\}\{2n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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334、 5、 已知 $\displaystyle F(x)=\int\_0^x (x-2t)f(t)\mathrm\{ d\} t$. (1)、 若 $\displaystyle f(x)$ 为偶函数, 证明 $\displaystyle F(x)$ 也是偶函数. (2)、 若 $\displaystyle f(x)$ 单调不增, 证明 $\displaystyle F(x)$ 单调不减. (北京邮电大学2023年数学分析考研试题) [定积分 ]
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https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、
\begin\{aligned\} F(x)=&\int\_0^\{-x\}(-x-2t)f(t)\mathrm\{ d\} t\stackrel\{t=-s\}\{=\}\int\_0^x (-x+2s)f(-s)(-\mathrm\{ d\} s)\\\\ =&\int\_0^x (x-2s)f(s)\mathrm\{ d\} s=F(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由题设, $\displaystyle f$ 递减, 而 $\displaystyle F(x)=x\int\_0^x f(t)\mathrm\{ d\} t-2\int\_0^x tf(t)\mathrm\{ d\} t$ 蕴含
\begin\{aligned\} F'(x)=&\int\_0^x f(t)\mathrm\{ d\} t+xf(x)-2xf(x) =\int\_0^x [f(t)-f(x)]\mathrm\{ d\} t\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle F$ 递增, 单调不减.跟锦数学微信公众号. [在线资料](
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---
335、 (6)、 设 $\displaystyle f(x)$ 连续,
\begin\{aligned\} g(x)=\frac\{1\}\{n!\}\int\_a^x (x-t)^n f(t)\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
计算 $\displaystyle g^\{(n+1)\}(x)$. (大连理工大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 注意被积区间和被积函数中都含有 $\displaystyle x$. 于是
\begin\{aligned\} &g'(x)=\frac\{1\}\{n!\}\int\_a^x n(x-t)^\{n-1\} f(t)\mathrm\{ d\} t,\\\\ &g''(x)=\frac\{1\}\{n!\}\int\_a^x n(n-1)(x-t)^\{n-2\} f(t)\mathrm\{ d\} t,\\\\ &\cdots, g^\{(n)\}(x)=\int\_a^x f(t)\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
336、 8、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,2]$ 上二阶连续可导, 且
\begin\{aligned\} \int\_0^2 f(x)\mathrm\{ d\} x=4, f(2)=1, f'(2)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle \int\_0^1 x^2f''(2x)\mathrm\{ d\} x$. (东南大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&\stackrel\{2x=t\}\{=\}\int\_0^2 \frac\{t^2\}\{4\}f''(t)\frac\{\mathrm\{ d\} t\}\{2\} =\frac\{1\}\{8\}\int\_0^2 t^2\mathrm\{ d\} f'(t) \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\frac\{1\}\{8\}\int\_0^2 2tf'(t)\mathrm\{ d\} t\\\\ =&-\frac\{1\}\{4\}\int\_0^2 t\mathrm\{ d\} f(t) \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\frac\{1\}\{4\}\left\[2-\int\_0^2 f(t)\mathrm\{ d\} t\right\]=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
337、 14、 设 $\displaystyle f: [a,b]\to\mathbb\{R\}$ 单调有界, 证明: $\displaystyle f$ 在 $\displaystyle [a,b]$ 上可积. (东南大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 不妨设 $\displaystyle f$ 递增. 对 $\displaystyle [a,b]$ 的任一分划 $\displaystyle T$, $\displaystyle f$ 在 $\displaystyle T$ 所属的每个小区间 $\displaystyle \Delta\_i$ 上的振幅为
\begin\{aligned\} \omega\_i=f(x\_i)-f(x\_\{i-1\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \sum\_\{i=1\}^n \omega\_i \Delta x\_i\leq \sum\_\{i=1\}^n [f(x\_i)-f(x\_\{i-1\})]\left\Vert T\right\Vert =[f(b)-f(a)]\left\Vert T\right\Vert . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta=\frac\{\varepsilon\}\{f(b)-f(a)+1\} > 0,\mathrm\{ s.t.\} \forall\ T: \left\Vert T\right\Vert < \delta, \sum\_T \omega\_i \Delta x\_i < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上可积.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
338、 (3)、 求定积分 $\displaystyle \int\_0^a x^2\sqrt\{a^2-x^2\}\mathrm\{ d\} x$, 其中 $\displaystyle a > 0$. (广西大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&\stackrel\{x=a\sin\theta\}\{=\}\int\_0^\frac\{\pi\}\{2\} a^2\sin^2\theta a\cos\theta\cdot a\cos\theta\mathrm\{ d\} \theta\\\\ &=\frac\{a^4\}\{4\}\int\_0^\frac\{\pi\}\{2\}\sin^22\theta\mathrm\{ d\} \theta=\left.\frac\{a^4\}\{8\}\left(\theta+\frac\{1\}\{4\}\sin4\theta\right)\right|\_0^\frac\{\pi\}\{2\}=\frac\{\pi a^4\}\{16\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
339、 (4)、 求由抛物线 $\displaystyle y^2=x$ 与直线 $\displaystyle x-2y-3=0$ 所围平面图形的面积. (广西大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 易知题中两曲线的交点为 $\displaystyle (1,-1), (9,3)$. 故所求
\begin\{aligned\} =\int\_\{-1\}^3 [(2y+3)-y^2]\mathrm\{ d\} y=\frac\{32\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
340、 9、 证明黎曼函数
\begin\{aligned\} R(x)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{1\}\{q\},&x\in(0,1)\mbox\{且\} x=\frac\{p\}\{q\}, (p,q)=1, p,q\in\mathbb\{Z\}\_+,\\\\ 0,&x=0,1\mbox\{或\} x\in[0,1]\backslash \mathbb\{Q\}\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle [0,1]$ 上黎曼可积. (哈尔滨工程大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \forall\ \varepsilon > 0, \eta > 0$, 又由 $\displaystyle [0,1]$ 中满足 $\displaystyle \frac\{1\}\{q\}\geq \varepsilon\Leftrightarrow 0 < q\leq\frac\{1\}\{\varepsilon\}$ 的有理数 $\displaystyle \frac\{p\}\{q\}$ 只有有限个 (设为 $\displaystyle m$ 个), 所以对 $\displaystyle [0,1]$ 的任意分划 $\displaystyle T$, $\displaystyle T$ 中包含这类点的小区间至多 $\displaystyle 2m$ 个, 在其上, $\displaystyle \omega\_\{k'\}\geq \varepsilon$. 因此, 当 $\displaystyle \left\Vert T\right\Vert < \frac\{\eta\}\{2m\}$ 时, 满足 $\displaystyle \omega\_\{k'\}\geq \varepsilon$ 的那些小区间的总长
\begin\{aligned\} \sum\_\{k'\}\Delta x\_\{k'\} \leq 2m\left\Vert T\right\Vert < \eta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由可积性准则 III 即知 $\displaystyle R(x)$ 在 $\displaystyle [0,1]$ 上可积.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
341、 5、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续且不恒为零, 同时 $\displaystyle \int\_a^b f(x)\mathrm\{ d\} x=0$. (1)、 证明 $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 上有一个零点; (2)、 若 $\displaystyle \int\_a^b xf(x)\mathrm\{ d\} x=0$, 证明 $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 上至少有两个零点. (哈尔滨工业大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle F(x)=\int\_a^x f(t)\mathrm\{ d\} t$, 则 $\displaystyle F(a)=F(b)=0$. 由 Rolle 定理知
\begin\{aligned\} \exists\ \xi\in (a,b),\mathrm\{ s.t.\} 0=F'(\xi)=f(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-1)、 若 $\displaystyle f(x)\geq 0, \forall\ x\in [a,b]$, 则
\begin\{aligned\} \int\_0^1 f(x)\mathrm\{ d\} x=0\Rightarrow f(x)\equiv 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
结论自明. 若 $\displaystyle f(x)\leq 0, \forall\ x\in [a,b]$, 则类似可得 $\displaystyle f\equiv 0$. (1-2)、 若
\begin\{aligned\} \exists\ x\_0\neq x\_1,\mathrm\{ s.t.\} f(x\_0) < 0 < f(x\_1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由连续函数介值定理知 $\displaystyle f$ 在 $\displaystyle (a,b)$ 上至少有一个零点 $\displaystyle \xi$. 往用反证法证明 $\displaystyle f$ 至少还有一个零点. 设 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上只有这个零点 $\displaystyle \xi$, 则
\begin\{aligned\} f(x) < 0, \forall\ x < \xi; f(x) > 0, \forall\ x > \xi,\qquad(\star) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
或
\begin\{aligned\} f(x) > 0, \forall\ x < \xi; f(x) < 0, \forall\ x > \xi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不妨设 $\displaystyle (\star)$ 成立, 则
\begin\{aligned\} &(x-\xi)f(x) > 0, \forall\ x\neq \xi\\\\ \Rightarrow&0 < \int\_a^b (x-\xi)f(x)\mathrm\{ d\} x=\int\_a^b xf(x)\mathrm\{ d\} x-\xi\int\_a^b f(x)\mathrm\{ d\} x=0\\\\ \Rightarrow&\mbox\{矛盾, 故有结论\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
342、 3、 设 $\displaystyle k$ 为正整数, 求 $\displaystyle \int\_0^\{k\pi\}|\cos x|\mathrm\{ d\} x$. (黑龙江大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=k\int\_0^\pi |\cos x|\mathrm\{ d\} x =2k\int\_0^\frac\{\pi\}\{2\}\cos x\mathrm\{ d\} x=2k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
343、 14、 求曲线 $\displaystyle r=1+\cos\theta$ 围成的图形的面积. (黑龙江大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 所求面积
\begin\{aligned\} =\frac\{1\}\{2\}\int\_0^\{2\pi\} r^2(\theta)\mathrm\{ d\} \theta=\frac\{3\pi\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
344、 2、 计算积分
\begin\{aligned\} \int\_0^1 \frac\{1\}\{x+\sqrt\{1-x^2\}\}\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华东师范大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&\stackrel\{x=\sin\theta\}\{=\}\int\_0^\frac\{\pi\}\{2\}\frac\{\cos\theta\}\{\sin \theta+\cos \theta\}\mathrm\{ d\} \theta\equiv I\\\\ &\stackrel\{\frac\{\pi\}\{2\}-\theta=s\}\{=\}\int\_0^\frac\{\pi\}\{2\}\frac\{\sin s\}\{\cos s+\sin s\}\mathrm\{ d\} s\equiv J\\\\ &=\frac\{1\}\{2\}(I+J) =\frac\{1\}\{2\}\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta=\frac\{\pi\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
345、 (2)、 求定积分 $\displaystyle \int\_0^\frac\{\pi\}\{4\}\frac\{x\mathrm\{ d\} x\}\{1+\cos 2x\}$. (华南师范大学2023年数学分析考研试题) [定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\int\_0^\frac\{\pi\}\{4\}\frac\{x\mathrm\{ d\} x\}\{2\cos^2x\} =\frac\{1\}\{2\}\int\_0^\frac\{\pi\}\{4\} x\sec^2x\mathrm\{ d\} x\\\\ =&\frac\{1\}\{2\}\int\_0^\frac\{\pi\}\{4\} x\mathrm\{ d\} \tan x =\frac\{1\}\{2\}\left(\frac\{\pi\}\{4\}-\int\_0^\frac\{\pi\}\{4\}\tan x\mathrm\{ d\} x\right)\\\\ =&\frac\{\pi\}\{8\}-\left.\frac\{1\}\{2\}(-\ln \cos x)\right|\_0^\frac\{\pi\}\{4\} =\frac\{\pi\}\{8\}-\frac\{1\}\{4\}\ln 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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发表于 2023-3-5 08:48:53
