Questions in category: 定积分 (Definite Integral)
分析 >> 数学分析 >> 定积分
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1. 求积分 $\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\mathrm{d}x$

Posted by haifeng on 2023-11-22 13:00:17 last update 2023-11-22 13:00:17 | Answers (1) | 收藏


证明:

\[\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\mathrm{d}x=\frac{22}{7}-\pi\]

 

 

参考:

知乎大佬是怎么"注意到"这么恐怖的积分的?_哔哩哔哩_bilibili

2. 计算 $I_n=e^{-1}\int_0^1 x^n e^x\mathrm{d}x$.

Posted by haifeng on 2023-10-28 19:58:58 last update 2023-10-28 20:37:58 | Answers (1) | 收藏


计算 $I_n=e^{-1}\int_0^1 x^n e^x\mathrm{d}x$ ($n=0,1,2,\ldots$)

 

容易证明

\[
I_n=\begin{cases}
1-nI_{n-1}, & n\geqslant 1,\\
1-e^{-1}, & n=0.
\end{cases}
\]

将递推公式改写为 $I_{n+1}=1-(n+1)*I_n$, 则使用 Sowya 可以计算如下:

首先计算 $I_0$:

 

>> setprecision(100)
Now the precision is: 100

------------------------

>> exp(1)
out> 2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274

 

令 e 等于上面的 exp(1)

>> e=2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274
 

>> 1-1/e
in> 1-1/2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274

out> 0.6321205588285576784044762298385391325541888689682321654921631983025385042551001966428527256540803562
------------------------

然后再计算 $I_1=1-I_0$:

>> 1-0.6321205588285576784044762298385391325541888689682321654921631983025385042551001966428527256540803562
in> 1-0.6321205588285576784044762298385391325541888689682321654921631983025385042551001966428527256540803562

out> 0.3678794411714423215955237701614608674458111310317678345078368016974614957448998033571472743459196438
------------------------


利用 printRecursiveSeries() 函数(Pro版本中提供)计算 $I_n$, $n=1,2,3,\ldots,20$.

>> printRecursiveSeries(1-(n+1)*I_n,I_n,0.3678794411714423215955237701614608674458111310317678345078368016974614957448998033571472743459196438,20,\n,linenumber)
[1]     0.3678794411714423215955237701614608674458111310317678345078368016974614957448998033571472743459196438
[2]     0.2642411176571153568089524596770782651083777379364643309843263966050770085102003932857054513081607124
[3]     0.2072766470286539295731426209687652046748667861906070070470208101847689744693988201428836460755178628
[4]     0.1708934118853842817074295161249391813005328552375719718119167592609241021224047194284654156979285488
[5]     0.1455329405730785914628524193753040934973357238121401409404162036953794893879764028576729215103572560
[6]     0.1268023565615284512228854837481754390159856571271591543575027778277230636721415828539624709378564640
[7]     0.1123835040693008414398016137627719268881004001098859194974805552059385542950089200222627034350047520
[8]     0.1009319674455932684815870898978245848951967991209126440201555583524915656399286398218983725199619840
[9]     0.0916122929896605836657161909195787359432288079117862038185999748275759092406422416029146473203421440
[10]    0.0838770701033941633428380908042126405677119208821379618140002517242409075935775839708535267965785600
[11]    0.0773522288626642032287810011536609537551688702964824200459972310333500164706465763206112052376358400
[12]    0.0717732536480295612546279861560685549379735564422109594480332275997998023522410841526655371483699200
[13]    0.0669477025756157036898361799711087858063437662512575271755680412026025694208659060153480170711910400
[14]    0.0627321639413801483422934804044769987111872724823946195420474231635640281078773157851277610033254400
[15]    0.0590175408792977748655977939328450193321909127640807068692886525465395783818402632230835849501184000
[16]    0.0557193459312356021504352970744796906849453957747086900913815592553667458905557884306626407981056000
[17]    0.0527711191689947634425999497338452583559282718299522684465134926587653198605515966787351064322048000
[18]    0.0501198549580942580332009047907853495932911070608591679627571321422242425100712597827680842203136000
[19]    0.0477227557962090973691828089750783577274689658436758087076144892977393923086460641274063998140416000
[20]    0.0455448840758180526163438204984328454506206831264838258477102140452121538270787174518720037191680000


------------------------

3. 证明: $\int_{0}^{\frac{\pi}{2}}x\cdot\sin^{2n-1}x\mathrm{d}\sin x=\frac{\pi}{4n}\cdot\biggl[1-\frac{(2n-1)!!}{(2n)!!}\biggr]$

Posted by haifeng on 2023-05-03 13:52:12 last update 2023-05-03 13:52:12 | Answers (0) | 收藏


证明: \[\int_{0}^{\frac{\pi}{2}}x\cdot\sin^{2n-1}x\mathrm{d}\sin x=\frac{\pi}{4n}\cdot\biggl[1-\frac{(2n-1)!!}{(2n)!!}\biggr]\]

 

参见问题2871的解答过程.

4. 令 $I(\alpha)=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{1}{1+\tan^{\alpha}x}\mathrm{d}x$, 求 $I'(\alpha)$.

Posted by haifeng on 2023-03-18 09:17:47 last update 2023-03-18 09:18:34 | Answers (1) | 收藏


令 $I(\alpha)=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{1}{1+\tan^{\alpha}x}\mathrm{d}x$, 求 $I'(\alpha)$.

5. 求定积分 $\int_0^{\pi}\frac{x}{\tan x}\mathrm{d}x$.

Posted by haifeng on 2022-12-15 15:03:25 last update 2022-12-15 15:08:26 | Answers (0) | 收藏


求定积分 $\int_0^{\pi}\frac{x}{\tan x}\mathrm{d}x$.

 

 

 


Rem: 题目来源, 在bilibili上看到的这道题目.

解这道定积分,花了我45分钟。你呢?_哔哩哔哩_bilibili

6. 设 $f(x)$ 在 $[0,1]$ 上连续且严格单调递增, $f(0)=0$. 证明: 对 $\forall\ x\in[0,1)$, 有 $e^{1-x}\int_0^x f(t)\mathrm{d}t < \int_0^1 f(x)\mathrm{d}x$.

Posted by haifeng on 2022-12-15 09:29:35 last update 2022-12-15 09:29:35 | Answers (1) | 收藏


设 $f(x)$ 在 $[0,1]$ 上连续且严格单调递增, $f(0)=0$. 证明: 对 $\forall\ x\in[0,1)$, 有

\[e^{1-x}\int_0^x f(t)\mathrm{d}t < \int_0^1 f(x)\mathrm{d}x.\]

7. 设 $f(x)\in C([0,2\pi])$ 是严格单调递增的, 证明 $\int_0^{2\pi}f(x)\sin x\mathrm{d}x < 0$.

Posted by haifeng on 2022-10-08 11:25:12 last update 2022-10-08 11:34:39 | Answers (1) | 收藏


若函数 $f(x)$ 在 $[0,2\pi]$ 上连续且严格单调递增, 问

\[I=\int_0^{2\pi}f(x)\sin x\mathrm{d}x\]

的符号是正还是负?

8. 求 $I=\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{n}{n^2+k^2+k}$.

Posted by haifeng on 2022-03-21 23:15:11 last update 2022-03-21 23:15:35 | Answers (1) | 收藏


利用定积分求极限

\[I=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{n}{n^2+k^2+k}.\]

9. 求 $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\csc^4 x\mathrm{d}x$

Posted by haifeng on 2022-03-16 21:51:22 last update 2022-03-16 22:03:10 | Answers (0) | 收藏


\[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\csc^4 x\mathrm{d}x\]

 

 

[Hint] 将被积函数拆成 $\csc^2 x\cdot\csc^2 x$, $\csc^2 x\mathrm{d}x=-\mathrm{d}\cot x$. 然后用分部积分.

 

10. 求定积分 $\int_0^{\frac{\pi}{2}}\frac{x\ln(\cos x)}{\tan x}\mathrm{d}x$.

Posted by haifeng on 2021-12-28 20:23:57 last update 2022-01-01 21:56:12 | Answers (3) | 收藏


求定积分

\[\int_0^{\frac{\pi}{2}}\frac{x\ln(\cos x)}{\tan x}\mathrm{d}x.\]

 

 


Remark:  题目来源于同事陶文清.

这道题的正确解答参见这里的第三个解答. 涉及多个知识点. 是很好的一道习题. 这个反常积分, 用通常的换元法、分部积分法都不好算. 甚至也暂时想不到是否能用复变函数的方法(即利用留数定理计算实变函数的积分). 

解答中需要用到无穷级数、常微分方程等.

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