## 张祖锦2023年数学专业真题分类70天之第16天
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346、 3、 (15 分) 判断积分 $\displaystyle \int\_0^\{2\pi\}\mathrm\{e\}^\{-x^2\}\cos x\mathrm\{ d\} x$ 的正负, 并给出理由. (华中科技大学2023年数学分析考研试题) [定积分 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} \mbox\{原式\}=&\int\_0^\pi +\int\_\pi^\{2\pi\}\mathrm\{e\}^\{-x^2\}\cos x\mathrm\{ d\} x\\\\ \stackrel\{x-\pi=t\}\{=\}&\int\_0^\pi \left\[\mathrm\{e\}^\{-x^2\}-\mathrm\{e\}^\{-(x+\pi)^2\}\right\]\cos x\mathrm\{ d\} x > 0! \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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347、 5、 (25 分) 求解下列问题. (1)、 证明: 方程 $\displaystyle (x+1)^\{x+1\}=\mathrm\{e\}\cdot x^x$ 只有唯一正实根. (2)、 若 $\displaystyle p(x)=\frac\{x^2-x\}\{2\}$, $\displaystyle f(x)$ 二阶可导, 则有
\begin\{aligned\} \int\_k^\{k+1\} f(x)\mathrm\{ d\} x =\frac\{f(k+1)+f(k)\}\{2\} -\int\_k^\{k+1\} f''(x)p(x-[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\}
其中 $\displaystyle [x]$ 为取整函数; (3)、 若 $\displaystyle \beta$ 为 (1) 中方程的正实根, 计算
\begin\{aligned\} \lim\_\{n\to\infty\}\left(\beta+\frac\{1\}\{n\}\right)\left(\beta+\frac\{2\}\{n\}\right)\cdots\left(\beta+\frac\{n\}\{n\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中科技大学2023年数学分析考研试题) [定积分 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle f(x)=(x+1)\ln(x+1)-1-x\ln x, x > 0$, 则 $\displaystyle f(0+0)=-1$,
\begin\{aligned\} \lim\_\{x\to+\infty\}f(x)=\lim\_\{x\to+\infty\}\left\[x\ln \left(1+\frac\{1\}\{x\}\right)-1+\ln (x+1)\right\] =+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由连续函数介值定理知 $\displaystyle f=0$ 在 $\displaystyle x > 0$ 至少有一个根. 又由 $\displaystyle f'(x)=\ln(x+1)-\ln x > 0$ 知 $\displaystyle f=0$ 有且仅有一个正根. (2)、
\begin\{aligned\} &\int\_k^\{k+1\} f''(x)p(x-[x])\mathrm\{ d\} x\\\\ =&\frac\{1\}\{2\} \int\_k^\{k+1\} f''(x)\left\[(x-k)^2-(x-k)\right\]\mathrm\{ d\} x\\\\ =&\frac\{1\}\{2\}\int\_k^\{k+1\} [(x-k)^2-(x-k)]\mathrm\{ d\} f'(x)\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}& -\frac\{1\}\{2\}\int\_k^\{k+1\} [2(x-k)-1]f'(x)\mathrm\{ d\} x\\\\ =&-\frac\{1\}\{2\} \int\_k^\{k+1\} [2(x-k)-1]\mathrm\{ d\} f(x)\\\\ =&-\frac\{1\}\{2\}[f(k+1)-(-1)f(k)]+\int\_k^\{k+1\} 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\}
(3)、 在第 2 步中令 $\displaystyle f(x)=\ln \left(\beta+\frac\{x\}\{n\}\right)$, 则
\begin\{aligned\} \int\_k^\{k+1\} \ln \left(\beta+\frac\{x\}\{n\}\right)\mathrm\{ d\} x=&\frac\{\ln \left(\beta+\frac\{k+1\}\{n\}\right)+\ln \left(\beta+\frac\{k\}\{n\}\right)\}\{2\}\\\\ &-\int\_k^\{k+1\} \frac\{-1\}\{n^2\left(\beta+\frac\{x\}\{n\}\right)^2\} p(x-[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\}
$\displaystyle k$ 从 $\displaystyle 0$ 到 $\displaystyle n-1$ 求和得
\begin\{aligned\} \int\_0^n \ln \left(\beta+\frac\{x\}\{n\}\right)\mathrm\{ d\} x =&\frac\{\ln \beta\}\{2\}+\sum\_\{k=1\}^\{n-1\}\ln \left(\beta+\frac\{k\}\{n\}\right) +\frac\{\ln(\beta+1)\}\{2\}+O\left(\frac\{1\}\{n^2\}\right)\\\\ =&\frac\{\ln \beta\}\{2\}+\sum\_\{k=1\}^n\ln \left(\beta+\frac\{k\}\{n\}\right) -\frac\{\ln(\beta+1)\}\{2\}+O\left(\frac\{1\}\{n^2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle n\to\infty$ 得
\begin\{aligned\} &\lim\_\{n\to\infty\}\sum\_\{k=1\}^n \ln \left(\beta+\frac\{k\}\{n\}\right)\\\\ =&\frac\{\ln (\beta+1)-\ln \beta\}\{2\} -\lim\_\{n\to\infty\}\int\_0^n \ln \left(\beta+\frac\{x\}\{n\}\right)\mathrm\{ d\} t\\\\ =&\ln \sqrt\{1+\frac\{1\}\{\beta\}\} -\lim\_\{n\to\infty\}n\int\_0^1 \ln (\beta+t)\mathrm\{ d\} t\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}&\ln \sqrt\{1+\frac\{1\}\{\beta\}\} -\lim\_\{n\to\infty\}n \left\[(\beta+1)\ln(\beta+1)-\beta\ln \beta-1\right\]\\\\ =&\ln \sqrt\{1+\frac\{1\}\{\beta\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lim\_\{n\to\infty\}\left(\beta+\frac\{1\}\{n\}\right)\left(\beta+\frac\{2\}\{n\}\right)\cdots\left(\beta+\frac\{n\}\{n\}\right) =\ln \sqrt\{1+\frac\{1\}\{\beta\}\}$.跟锦数学微信公众号. [在线资料](
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348、 (7)、 求定积分 $\displaystyle \int\_0^1 \frac\{x\}\{\mathrm\{e\}^x+\mathrm\{e\}^\{1-x\}\}\mathrm\{ d\} x$. (吉林大学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](
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\begin\{aligned\} \mbox\{原式\}&\stackrel\{1-x=t\}\{=\}\int\_0^1\frac\{1-t\}\{\mathrm\{e\}^\{1-t\}+\mathrm\{e\}^t\}\mathrm\{ d\} t\equiv J\\\\ &=\frac\{1\}\{2\}\left(\mbox\{原式\}+J\right)=\frac\{1\}\{2\}\int\_0^1 \frac\{1\}\{\mathrm\{e\}^\{1-t\}+\mathrm\{e\}^t\}\mathrm\{ d\} t =\frac\{1\}\{2\}\int\_0^1 \frac\{\mathrm\{e\}^t\}\{\mathrm\{e\}+\mathrm\{e\}^\{2t\}\}\mathrm\{ d\} t\\\\ &\stackrel\{\mathrm\{e\}^t=s\}\{=\}\frac\{1\}\{2\}\int\_1^\{\mathrm\{e\}\}\frac\{\mathrm\{ d\} s\}\{\mathrm\{e\}+s^2\} \stackrel\{s=\sqrt\{\mathrm\{e\}\}t\}\{=\}\frac\{1\}\{2\}\int\_\{\frac\{1\}\{\sqrt\{\mathrm\{e\}\}\}\}^\{\sqrt\{\mathrm\{e\}\}\} \frac\{\sqrt\{\mathrm\{e\}\}\mathrm\{ d\} t\}\{\mathrm\{e\}(1+t^2)\}\\\\ &=\frac\{1\}\{2\sqrt\{\mathrm\{e\}\}\}\left(\arctan \sqrt\{\mathrm\{e\}\}-\arctan \frac\{1\}\{\sqrt\{\mathrm\{e\}\}\}\right). \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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349、 3、 解答题. (0-22)、 求圆 $\displaystyle x^2+y^2=2y$ 内位于抛物线 $\displaystyle y=x^2$ 下方部分的面积. (吉林师范大学2023年(学科数学)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\} \xlongequal\{\tiny\mbox\{对称性\}\}&2\int\_0^1 \left\[x^2-\left\[1-\sqrt\{1-x^2\}\right\]\right\]\mathrm\{ d\} x\\\\ \stackrel\{x=\sin\theta\}\{=\}&2\left(\frac\{1\}\{3\}-1+\frac\{\pi\}\{4\}\right)=\frac\{\pi\}\{2\}-\frac\{4\}\{3\}. \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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350、 (6)、 (可能有误) 求积分 $\displaystyle \int\_\{-1\}^1 \frac\{|x|^\frac\{1\}\{3\}\left(\sin \sqrt[3]\{x\}\cos \sqrt[3]\{x\}+\sin x\cos x\right)\}\{1+x^\frac\{4\}\{3\}\}\mathrm\{ d\} x$. (暨南大学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) / 被积函数关于 $\displaystyle x$ 是奇函数, 而积分为 $\displaystyle 0$.跟锦数学微信公众号. [在线资料](
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351、 (3)、 设 $\displaystyle f(x)=\int\_1^x \mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t$, 求 $\displaystyle \int\_0^1 x^2f(x)\mathrm\{ d\} x$. (南昌大学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/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\int\_0^1 f(x)\mathrm\{ d\} \frac\{x^3\}\{3\} \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} f(1)\frac\{1\}\{3\}-\int\_0^1 f'(x)\frac\{x^3\}\{3\}\mathrm\{ d\} x\\\\ =&-\frac\{1\}\{3\}\int\_0^1 x^3\mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x\stackrel\{x^2=t\}\{=\}\cdots\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \frac\{2-\mathrm\{e\}\}\{6\mathrm\{e\}\}. \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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352、 5、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上有界, $\displaystyle \left\\{c\_n\right\\}\subset [a,b]$, 数列 $\displaystyle \left\\{c\_n\right\\}$ 收敛. 证明: 若 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上只有 $\displaystyle c\_n\ (n=1,2,\cdots)$ 为其间断点, 则 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上可积. (南昌大学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) / 题中函数 $\displaystyle f$ 的不连续点只有一个聚点, 就是 $\displaystyle \left\\{c\_n\right\\}$ 的极限. 我们证明一般的结论: 若有界函数的不连续点集只有有限多个聚点, 则 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上黎曼可积. 具体来说: 设函数 $\displaystyle f(x)$ 在区间 $\displaystyle [a,b]$ 上有界, 且它的不连续点集只有有限个聚点, 则 $\displaystyle f(x)\in R[a,b]$. 事实上, 设 $\displaystyle f$ 的不连续点集 $\displaystyle D$ 的聚点为 $\displaystyle \left\\{x\_1,\cdots,x\_r\right\\}$, 则对 $\displaystyle \forall\ \varepsilon > 0, \eta > 0$, 取总长度小于 $\displaystyle \frac\{\varepsilon\}\{2\}$ 的 $\displaystyle r$ 个小区间 $\displaystyle I\_i\left(i=1,\cdots,r\right)$ 盖住个 $\displaystyle x\_i$. 则 $\displaystyle [a,b] \backslash \bigcup\_\{i=1\}^r I\_i$ 只有 $\displaystyle f$ 的有限多个不连续点, 又找总长度小于 $\displaystyle \frac\{\varepsilon\}\{2\}$ 的有限多个小区间 $\displaystyle \left\\{J\_j\right\\}\_\{j=1\}^s$ 盖住这些 $\displaystyle f$ 的不连续点. 如此, $\displaystyle f$ 在 $\displaystyle [a,b]\backslash \bigcup\_\{i=1\}^r I\_i\backslash \bigcup\_\{j=1\}^s J\_j$ 上连续, 而一致连续, 可加上有限多个适当分点 $\displaystyle \left\\{x\_k\right\\}$, 使得每个小区间的振幅都 $\displaystyle < \eta$. 将 $\displaystyle I\_i,J\_j$ 的端点并上 $\displaystyle \left\\{x\_k\right\\}$ 得到分划 $\displaystyle T$, 则 $\displaystyle T$ 中振幅 $\displaystyle \geq \eta$ 的区间总长
\begin\{aligned\} \leq \sum\_i |I\_i|+\sum\_j|I\_j| < \frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据可积的第三充分必要条件即使 $\displaystyle f\in R[a,b]$.跟锦数学微信公众号. [在线资料](
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353、 8、 (10 分) 比较 $\displaystyle \int\_0^\pi \mathrm\{e\}^\{\cos^2x\}\mathrm\{ d\} x$ 与 $\displaystyle \frac\{3\}\{2\}\pi$ 的大小. (南京大学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) /
\begin\{aligned\} \int\_0^\pi \mathrm\{e\}^\{\cos^2x\}\mathrm\{ d\} x > \int\_0^\pi (1+\cos^2x)\mathrm\{ d\} x=\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\}
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354、 (2)、 求积分 $\displaystyle \int\_0^\frac\{\pi\}\{2\}\frac\{1\}\{1+\tan^\{2023\}x\}\mathrm\{ d\} x$. (南京航空航天大学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\}\{2\}\frac\{\cos^\{2023\}x\}\{\sin^\{2023\}x+\cos^\{2023\}x\}\mathrm\{ d\} x \stackrel\{\frac\{\pi\}\{2\}-x=t\}\{=\}\int\_0^\frac\{\pi\}\{2\}\frac\{\sin^\{2023\}t\}\{\sin^\{2023\}t+\cos^\{2023\}t\}\mathrm\{ d\} t\equiv I\\\\ =&\frac\{1\}\{2\}\left(\mbox\{原式\}+I\right)=\frac\{1\}\{2\}\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} t=\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\}
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355、 (3)、 (20 分) 设函数 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上可积, 且对 $\displaystyle [a,b]$ 上的任一连续函数 $\displaystyle g(x)$, 有 $\displaystyle \int\_a^b f(x)g(x)\mathrm\{ d\} x=0$. 证明: $\displaystyle f(x)$ 在连续点的取值为 $\displaystyle 0$. (山东大学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) / 用反证法. 若 $\displaystyle f$ 在其连续点 $\displaystyle x\_0$ 处不为 $\displaystyle 0$, 则不妨设 $\displaystyle x\_0\in (a,b)$ (端点处类似处理), 也不妨设 $\displaystyle f(x\_0) > 0$ (若不然考虑 $\displaystyle -f$ 即可). 由 $\displaystyle \lim\_\{x\to x\_0\} f(x)=f(x\_0)$ 知 $\displaystyle \exists\ \delta\in \left(0,\min\left\\{x\_0-a,b-x\_0\right\\}\right),\mathrm\{ s.t.\}$
\begin\{aligned\} \forall\ x\in U(x\_0;\delta), |f(x)-f(x\_0)| < \frac\{f(x\_0)\}\{2\}\Rightarrow f(x) > \frac\{f(x\_0)\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle g(x)=\left\\{\begin\{array\}\{llllllllllll\}1,&x\in U\left(x\_0;\frac\{\delta\}\{2\}\right),\\\\ 0,&x\in \left\[a,x\_0-\delta\right\]\cup \left\[x\_0+\delta,b\right\],\\\\ \mbox\{线性连接\},&\mbox\{其他\},\end\{array\}\right.$ 则
\begin\{aligned\} 0&\xlongequal\{\tiny\mbox\{题设\}\} \int\_a^b f(x)g(x)\mathrm\{ d\} x=\int\_\{U(x\_0;\delta)\}f(x)g(x)\mathrm\{ d\} x\\\\ &\geq \int\_\{U\left(x\_0;\frac\{\delta\}\{2\}\right)\}f(x)g(x)\mathrm\{ d\} x \geq \frac\{f(x\_0)\}\{2\}\cdot 1\cdot \delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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356、 (3)、 求抛物线 $\displaystyle y^2=x$ 与直线 $\displaystyle x-2y-3=0$ 所围区域的面积. (陕西师范大学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) / 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\}
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---
357、 8、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,a]$ $\displaystyle (a > 1)$ 上连续. 证明:
\begin\{aligned\} \int\_\frac\{1\}\{a\}^a f(x)\mathrm\{ d\} x=\frac\{1\}\{2\}\int\_\frac\{1\}\{a\}^a \left\[f(x)+\frac\{1\}\{x^2\}f\left(\frac\{1\}\{x\}\right)\right\]\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\}
(陕西师范大学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\} &\int\_\frac\{1\}\{a\}^a f(x)\mathrm\{ d\} x \stackrel\{x=\frac\{1\}\{t\}\}\{=\}\int\_a^\frac\{1\}\{a\} f\left(\frac\{1\}\{t\}\right)\cdot \left(-\frac\{1\}\{t^2\}\right)\mathrm\{ d\} t\\\\ =&\frac\{1\}\{2\}\int\_\frac\{1\}\{a\}^a \left\[f(x)+\frac\{1\}\{x^2\}f\left(\frac\{1\}\{x\}\right)\right\]\mathrm\{ d\} x\left(\mbox\{上两式的算术平均\}\right). \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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---
358、 10、 已知 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 且 $\displaystyle I=\int\_0^1 f(x)\mathrm\{ d\} x\neq 0$. 证明: 存在 $\displaystyle x\_1,x\_2\in (0,1), x\_1\neq x\_2$, 使得
\begin\{aligned\} \frac\{1\}\{f(x\_1)\}+\frac\{1\}\{f(x\_2)\}=\frac\{2\}\{I\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(上海财经大学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) / 设 $\displaystyle F(x)=\int\_0^x f(t)\mathrm\{ d\} t$, 则 $\displaystyle F(0)=0, F(1)=I$. 据连续函数介值定理,
\begin\{aligned\} \exists\ \xi\in (0,1),\mathrm\{ s.t.\} F(\xi)=\frac\{I\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由积分中值定理,
\begin\{aligned\} \exists\ 0 < x\_1 < \xi,\mathrm\{ s.t.\} \frac\{I\}\{2\}&=F(\xi)=\int\_0^\xi f(t)\mathrm\{ d\} t=f(x\_1)\xi,\\\\ \exists\ \xi < x\_2 < 1,\mathrm\{ s.t.\} \frac\{I\}\{2\}&=F(1)-F(\xi)=\int\_\xi^1 f(t)\mathrm\{ d\} t =f(x\_2)(1-\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} \frac\{1\}\{f(x\_1)\}+\frac\{1\}\{f(x\_2)\}=\frac\{2\xi\}\{I\}+\frac\{2(1-\xi)\}\{I\} =\frac\{2\}\{I\}. \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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---
359、 9、 解答如下问题: (1)、 若 $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 上连续且对任意的 $\displaystyle [\alpha,\beta]\subset (a,b)$, 有 $\displaystyle \int\_\alpha^\beta f(x)\mathrm\{ d\} x=0$, 证明: $\displaystyle f(x)\equiv 0, \forall\ x\in (a,b)$; (上海大学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) / 对 $\displaystyle \forall\ x\in (a,b)$, $\displaystyle \forall\ h\in \left(0,\min\left\\{x-a,b-x\right\\}\right)$,
\begin\{aligned\} \int\_x^\{x+h\} f(t)\mathrm\{ d\} t=0\Rightarrow f(x)=\lim\_\{h\to 0\}\frac\{1\}\{h\}\int\_x^\{x+h\}f(t)\mathrm\{ d\} t=0. \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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---
360、 (2)、 若 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, $\displaystyle \int\_a^b f(x)\mathrm\{ d\} x=0$ 且 $\displaystyle \int\_a^b xf(x)\mathrm\{ d\} x=0$, 证明: 至少存在两个不同的点 $\displaystyle x\_1,x\_2\in (a,b)$, 使得 $\displaystyle f(x\_1)=f(x\_2)=0$. (上海大学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) / 设 $\displaystyle F(x)=\int\_a^x f(t)\mathrm\{ d\} t$, 则 $\displaystyle F(a)=F(b)=0$,
\begin\{aligned\} 0=\int\_a^b xf(x)\mathrm\{ d\} x=\int\_a^b x\mathrm\{ d\} F(x) \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\int\_a^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\}
再设 $\displaystyle G(x)=\int\_a^x F(t)\mathrm\{ d\} t$, 则
\begin\{aligned\} &G(a)=G(b)=0, G'(a)=G'(b)=0\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ a < \eta < b,\mathrm\{ s.t.\} G'(\eta)=0=G'(a)=G'(b)\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ a < x\_1 < \eta < x\_2 < b,\mathrm\{ s.t.\} f(x\_1)=G''(x\_1)=G''(x\_2)=f(x\_2). \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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---
361、 (2)、 Riemann 可积函数一定存在原函数. (上海交通大学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) / $\displaystyle \times$. 比如 $\displaystyle f(x)=[x], x\in [0,1]$ Riemann 可积, 但因为由第一类间断点知 $\displaystyle f$ 一定没有原函数. 若不然, 与导数极限定理矛盾.跟锦数学微信公众号. [在线资料](
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---
362、 (4)、 函数 $\displaystyle |f|$ 在 $\displaystyle [a,b]$ 上 Riemann 可积, 则函数 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上也 Riemann 可积. (上海交通大学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) / $\displaystyle \times$. 比如 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}1,&x\in [a,b]\cap \mathbb\{Q\},\\\\ -1,&x\in [a,b]\backslash \mathbb\{Q\}.\end\{array\}\right.$跟锦数学微信公众号. [在线资料](
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---
363、 10、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上二阶连续可微, 且 $\displaystyle f'(0)=0, f''(0)\neq 0$. 证明: 对任意的 $\displaystyle x\in (0,1)$, 存在 $\displaystyle \xi(x)\in (0,x)$, 使得
\begin\{aligned\} \int\_0^x f(t)\mathrm\{ d\} t=f\left(\xi(x)\right)x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且 $\displaystyle \lim\_\{x\to 0^+\}\frac\{\xi(x)\}\{x\}=\frac\{1\}\{\sqrt\{3\}\}$. (苏州大学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) / 设 $\displaystyle F(x)=\int\_0^x f(t)\mathrm\{ d\} t$, 则
\begin\{aligned\} \int\_0^x f(t)\mathrm\{ d\} t=F(x)-F(0)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} F'\left(\xi(x)\right)x=f\left(\xi(x)\right)x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle 0 < \xi(x) < x$. 由函数极限的保号性知若 $\displaystyle \lim\_\{x\to 0^+\}\frac\{\xi(x)\}\{x\}$ 存在, 则它 $\displaystyle \geq 0$. 往证明这个极限存在, 并求出它的值. 据 Taylor 展式知
\begin\{aligned\} \int\_0^x f(t)\mathrm\{ d\} t=&F(x)=F(0)+F'(0)x+\frac\{F''(0)\}\{2\}x^2 +\frac\{F'''(0)\}\{6\}x^3+o(x^3)\\\\ =&f(0)x+\frac\{f''(0)\}\{6\}x^3+o(x^3),\\\\ f\left(\xi(x)\right)x=&x\left\[f(0)+f'(0)\xi(x)+\frac\{f''(0)\}\{2\}\xi^2(x)+o(x^2)\right\]\\\\ =&f(0)x+\frac\{f''(0)\}\{2\}\xi^2(x)x+o(x^3). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据题中等式知
\begin\{aligned\} &\frac\{f''(0)\}\{6\}x^3+o(x^3)=\frac\{f''(0)\}\{2\}\xi^2(x)x+o(x^3)\\\\ \Rightarrow&\frac\{f''(0)\}\{6\}+o(1)=\frac\{f''(0)\}\{2\}\left\[\frac\{\xi(x)\}\{x\}\right\]^2+o(1)\\\\ \stackrel\{x\to 0^+\}\{\Rightarrow\}&\lim\_\{x\to 0^+\}\left\[\frac\{\xi(x)\}\{x\}\right\]^2=\frac\{1\}\{3\} \Rightarrow \lim\_\{x\to 0^+\}\frac\{\xi(x)\}\{x\}=\frac\{1\}\{\sqrt\{3\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意最后一步我们用了如下用定义可以直接验证的结论: $\displaystyle \lim\_\{x\to x\_0\} g(x)=0, g(x)\geq 0\Rightarrow \lim\_\{x\to x\_0\} \sqrt\{g(x)\}=0$.跟锦数学微信公众号. [在线资料](
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---
364、 3、 求定积分 $\displaystyle \int\_0^1 \frac\{\ln(1+x)\}\{1+x^2\}\mathrm\{ d\} x$. (太原理工大学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) /
\begin\{aligned\} \mbox\{原式\}&\stackrel\{x=\tan\theta\}\{=\}\int\_0^\frac\{\pi\}\{4\}\ln (1+\tan \theta)\mathrm\{ d\} \theta =\int\_0^\frac\{\pi\}\{4\}\ln(\sin\theta+\cos\theta)\mathrm\{ d\} \theta-\int\_0^\frac\{\pi\}\{4\}\ln \cos\theta\mathrm\{ d\} \theta\\\\ &=\int\_0^\frac\{\pi\}\{4\}\ln \left\[\sqrt\{2\}\sin\left(\theta+\frac\{\pi\}\{4\}\right)\right\]\mathrm\{ d\} \theta -\int\_0^\frac\{\pi\}\{4\}\ln\cos\theta\mathrm\{ d\} \theta\\\\ &\stackrel\{\theta+\frac\{\pi\}\{4\}=\tau\}\{=\}\frac\{1\}\{2\}\ln 2\cdot\frac\{\pi\}\{4\} +\int\_\frac\{\pi\}\{4\}^\frac\{\pi\}\{2\}\ln \sin\tau\mathrm\{ d\} \tau -\int\_0^\frac\{\pi\}\{4\}\ln \cos\theta\mathrm\{ d\} \theta\\\\ &\stackrel\{\frac\{\pi\}\{2\}-\theta=\tau\}\{=\}\frac\{\pi\}\{8\}\ln 2+\int\_\frac\{\pi\}\{4\}^\frac\{\pi\}\{2\}\ln \sin\tau\mathrm\{ d\} \tau -\int\_\frac\{\pi\}\{4\}^\frac\{\pi\}\{2\}\ln \sin\tau\mathrm\{ d\} \tau=\frac\{\pi\}\{8\}\ln 2. \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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---
365、 6、 曲线 $\displaystyle y=ax^2\ (a > 0, x\geq 0)$ 与 $\displaystyle y=1-x^2$ 交于点 $\displaystyle A$, 过原点 $\displaystyle O$ 和点 $\displaystyle A$ 的直线与 $\displaystyle y=ax^2$ 围成平面区域. 问 $\displaystyle a$ 为何值时, 该平面图形绕 $\displaystyle x$ 轴旋转一周所得旋转体体积最大? (太原理工大学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) / 容易求得 $\displaystyle A\left(\frac\{1\}\{\sqrt\{1+a\}\},\frac\{a\}\{1+a\}\right)$. 而过原点 $\displaystyle O$ 和点 $\displaystyle A$ 的直线为 $\displaystyle y=\frac\{a\}\{\sqrt\{1+a\}\}x$. 故旋转体的体积
\begin\{aligned\} V(a)=&\int\_0^\frac\{1\}\{\sqrt\{1+a\}\} \pi\left\[\frac\{a^2x^2\}\{1+a\}-a^2x^4\right\]\mathrm\{ d\} x =\frac\{2\pi a^2\}\{15(1+a)^\frac\{5\}\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle V'(a)=\frac\{\pi a(4-a)\}\{15(1+a)^\frac\{7\}\{2\}\}$ 知 $\displaystyle \max V=V(4)=\frac\{32\pi\}\{375\sqrt\{5\}\}$.跟锦数学微信公众号. [在线资料](
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---
366、 2、 计算题. (1)、 求 $\displaystyle \int\_0^\pi\frac\{\mathrm\{ d\} x\}\{3+\sin^2x\}$. (武汉大学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-\frac\{\pi\}\{2\}=t\}\{=\}\int\_\{-\frac\{\pi\}\{2\}\}^\frac\{\pi\}\{2\} \frac\{\mathrm\{ d\} t\}\{3+\cos^2t\} \xlongequal\{\tiny\mbox\{对称性\}\} 2\int\_0^\frac\{\pi\}\{2\}\frac\{\mathrm\{ d\} t\}\{3+\cos^2t\}\\\\ &=2\int\_0^\frac\{\pi\}\{2\}\frac\{\sin^2t+\cos^2t\}\{4\cos^2t+3\sin^2t\}\mathrm\{ d\} t =2\int\_0^\frac\{\pi\}\{2\}\frac\{\mathrm\{ d\} \tan t\}\{4+3\tan^2t\}\\\\ &\stackrel\{\sqrt\{3\}\tan t=2s\}\{=\}2\int\_0^\infty \frac\{\frac\{2\}\{\sqrt\{3\}\}\mathrm\{ d\} s\}\{4(1+s^2)\} =\frac\{\pi\}\{2\sqrt\{3\}\}. \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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---
367、 (3)、 问是否存在 $\displaystyle [0,5]$ 上的函数 $\displaystyle f(x)$ 满足下列条件, 并说明理由. (3-1)、 $\displaystyle f(x)$ 连续可微; (3-2)、 $\displaystyle f(0)=f(5)=1$; (3-3)、 $\displaystyle |f'(x)|\leq\frac\{2\}\{5\}$; (3-4)、 $\displaystyle \left|\int\_0^5 f(x)\mathrm\{ d\} x\right|\leq \frac\{5\}\{2\}$. (武汉大学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) / 不存在. 用反证法. 若存在满足题设的 $\displaystyle f$, 则当 $\displaystyle 0\leq x\leq \frac\{5\}\{2\}$ 时,
\begin\{aligned\} f(x)=f(x)-f(0)+f(0)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f'(\xi\_x)x+1\geq -\frac\{2\}\{5\}x+1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle \frac\{5\}\{2\}\leq x\leq 1$ 时,
\begin\{aligned\} f(x)=&f(x)-f(5)+f(5)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f'(\eta\_x)(x-5)+1\\\\ \geq& \frac\{2\}\{5\}(x-5)+1 =\frac\{2\}\{5\}x-1\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
\begin\{aligned\} &\frac\{5\}\{2\}\geq \left|\int\_0^5 f(x)\mathrm\{ d\} x\right| =\int\_0^5 f(x)\mathrm\{ d\} x\\\\ \geq& \int\_0^\frac\{5\}\{2\}\left(1-\frac\{2\}\{5\}x\right)\mathrm\{ d\} x +\int\_\frac\{5\}\{2\}^5 \left(\frac\{2\}\{5\}x-1\right)\mathrm\{ d\} x=\frac\{5\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}1-\frac\{2\}\{5\}x,&0\leq x\leq\frac\{5\}\{2\}\\\\ \frac\{2\}\{5\}x-1,&\frac\{5\}\{2\}\leq x\leq 5\end\{array\}\right.\Rightarrow f'\left(\frac\{5\}\{2\}\right)=\left\\{\begin\{array\}\{llllllllllll\}f'\_+\left(\frac\{5\}\{2\}\right)=\frac\{2\}\{5\},\\\\ f'\_-\left(\frac\{5\}\{2\}\right)=-\frac\{2\}\{5\}.\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\}
这是一个矛盾. 故有结论.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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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368、 7、 (15 分) 设 $\displaystyle f(x)$ 是 $\displaystyle \mathbb\{R\}$ 上的连续可微函数, 且
\begin\{aligned\} f(x)=x+x\int\_0^1 f(t)\mathrm\{ d\} t+x^2\lim\_\{x\to 0\}\frac\{f(x)\}\{x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle f(x)$. (西南大学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) / 设 $\displaystyle A=\int\_0^1 f(t)\mathrm\{ d\} t, B=\lim\_\{x\to 0\}\frac\{f(x)\}\{x\}$, 则
\begin\{aligned\} f(x)=x+Ax+Bx^2\Rightarrow B=\lim\_\{x\to 0\}\frac\{f(x)\}\{x\}=A+1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上式的一个式子对 $\displaystyle x$ 在 $\displaystyle [0,1]$ 上积分得
\begin\{aligned\} A=(A+1)\frac\{1\}\{2\}+B\frac\{1\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
解得 $\displaystyle A=5, B=6$, $\displaystyle f(x)=6x(1+x)$.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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 08:49:26
