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分析 >> 数学分析 >> 定积分
Questions in category: 定积分 (Definite Integral).

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1

$\int_0^1 |f(x)-g(x)|dx\leqslant\frac{1}{2}$

Posted by haifeng on 2017-09-26 19:16:45 last update 2017-09-26 19:16:45 | Answers (0) | 收藏

设 $f(x)$ 和 $g(x)$ 是 $[0,1]$ 区间上的单调递增函数, 满足 $0\leqslant f(x), g(x)\leqslant 1$, 且 $\int_0^1 f(x)dx=\int_0^1 g(x)dx$, 证明

\[
\int_0^1 |f(x)-g(x)|dx\leqslant\frac{1}{2}.
\]

2

设 $f(x)$ 是 $[0,1]$ 上的二次连续可微函数, 满足 $f(0)=f(1)=0$, $f'(0)=1$, $f'(1)=0$. 证明 $\int_0^1 (f''(x))^2 dx\geqslant 4$.

Posted by haifeng on 2017-06-23 18:08:42 last update 2017-06-23 18:14:08 | Answers (1) | 收藏

设 $f(x)$ 是 $[0,1]$ 上的二次连续可微函数, 满足 $f(0)=f(1)=0$, $f'(0)=1$, $f'(1)=0$. 证明

\[
\int_0^1 (f''(x))^2 dx\geqslant 4.
\]

 

 


相关问题:

问题1473

问题1911  问题1912  问题1913  问题1863

3

设 $f\in C^2[a,b]$, 且 $f(a)=f'(a)=f'(b)=0$, 记 $M=\max_{x\in[a,b]}|f(x)|$. 证明下面的不等式.

Posted by haifeng on 2017-03-14 09:08:52 last update 2017-03-14 09:09:54 | Answers (0) | 收藏

设 $f\in C^2[a,b]$, 且 $f(a)=f'(a)=f'(b)=0$, 记 $M=\max_{x\in[a,b]}|f(x)|$. 则有

\[
M^2\leqslant\frac{(b-a)^3}{\pi}\int_a^b |f''(x)|^2 dx.
\]

 


[hint]

利用 Wirtinger 不等式 以及问题1912的结论.

4

设 $f\in C^1[a,b]$, 且 $f(a)=0$, 讨论 $f$ 与 $f'(x)$ 的关系.

Posted by haifeng on 2017-03-14 08:52:58 last update 2017-03-14 09:56:48 | Answers (1) | 收藏

设 $f\in C^1[a,b]$, 且 $f(a)=0$, 证明

\[
\sup_{x\in[a,b]}|f(x)|\leqslant\sqrt{b-a}\cdot\sqrt{\int_a^b (f'(x))^2 dx}.
\]

 

特别的, 如果 $a=0, b=1$, 则有 $\sup_{x\in[0,1]}|f(x)|\leqslant\sqrt{\int_0^1 (f'(x))^2 dx}.$ (见问题1863)


References:

Paulo Ney de Souza, Jorge-Nuno Silva, Berkeley Problems in Mathematics,Third Edition.

5

Wirtinger 不等式(Wirtinger's inequality)

Posted by haifeng on 2017-03-14 08:38:32 last update 2017-03-14 09:53:26 | Answers (0) | 收藏

假设 $g\in C^1[a,b]$, 且 $g(a)=g(b)=0$. 则有

\[
\int_a^b g^2(t)dt\leqslant\Bigl(\frac{b-a}{\pi}\Bigr)^2\int_a^b|g'(t)|^2 dt.
\]

 

References:

https://arxiv.org/abs/1407.6871
https://en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions

Wirtinger's inequality is seen as the one-dimensional version of Friedrichs' inequality.

6

求 $\int_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx$.

Posted by haifeng on 2017-03-13 18:02:41 last update 2017-03-14 10:40:03 | Answers (1) | 收藏

求 \[\int_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx.\]

 


[相关的积分]

证明:

\[
\int_0^1 \frac{1}{\sqrt{1-x^4}}dx\cdot\int_0^1 \frac{x^2}{\sqrt{1-x^4}}dx=\frac{\pi}{4}.
\]

 

如果我们记 $f(x)=\frac{1}{\sqrt{1-x^4}}$, $g(x)=\frac{x^2}{\sqrt{1-x^4}}$, 则

\[
\begin{aligned}
​f(x)+g(x)=\frac{1+x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1+x^2}{1-x^2}},\\
f(x)-g(x)=\frac{1-x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1-x^2}{1+x^2}},\\
​\end{aligned}
\]

于是 $(f(x)+g(x))(f(x)-g(x))=1$, 因此原积分

\[
​\begin{split}
\int_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx&=\int_0^1\sqrt{\frac{1-x^2}{1+x^2}}\\
​&=\int_0^1(f(x)-g(x))dx\\
&=\int_0^1 \frac{1}{\sqrt{1-x^4}}dx-\int_0^1 \frac{x^2}{\sqrt{1-x^4}}dx
​\end{split}
\]

 

References:

吉米多维奇, 《数学分析习题集题解》(五) 3872.

 


[分析]

设 $x^2=\sin\theta$, $\theta\in[0,\frac{\pi}{2}]$, 则 $x=\sqrt{\sin\theta}$, $dx=\frac{1}{2\sqrt{\sin\theta}}\cos\theta d\theta$.

\[
\begin{split}
\text{原式}&=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\theta}}{1+\sin\theta}\cdot\frac{\cos\theta}{2\sqrt{\sin\theta}}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2\theta}{2\sqrt{\sin\theta}(1+\sin\theta)}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\sin^2\theta}{2\sqrt{\sin\theta}(1+\sin\theta)}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\sin\theta}{2\sqrt{\sin\theta}}d\theta\\
&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin\theta}}d\theta-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\theta}d\theta,
\end{split}
\]

其中积分

\[
\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\theta}d\theta\stackrel{t=\sqrt{\sin\theta}}{=}2\int_0^1\frac{t^2}{\sqrt{1-t^4}}dt
\]

是一个椭圆积分, 可以参考问题998 .

7

设 $I(x)=\int_{\pi}^{2\pi}\frac{y\sin(xy)}{y-\sin y}dy$, 求 $\int_{0}^{1}I(x)dx$.

Posted by haifeng on 2016-12-27 07:07:47 last update 2016-12-27 07:07:47 | Answers (1) | 收藏

设 $I(x)=\int_{\pi}^{2\pi}\frac{y\sin(xy)}{y-\sin y}dy$, 求 $\int_{0}^{1}I(x)dx$.

8

计算定积分 $\int_0^1 \frac{x^2\arcsin x}{\sqrt{1-x^2}}dx$

Posted by haifeng on 2016-08-23 14:54:56 last update 2016-08-23 14:54:56 | Answers (1) | 收藏

计算定积分

\[
\int_0^1 \frac{x^2\arcsin x}{\sqrt{1-x^2}}dx
\]

9

求定积分 $\int_1^2 \frac{1}{x(1+x^n)}dx$.

Posted by haifeng on 2016-04-29 04:51:05 last update 2016-04-29 04:51:05 | Answers (1) | 收藏

\[
\int_1^2 \frac{1}{x(1+x^n)}dx
\]

10

求定积分 $\int_1^3 \sqrt{(3-x)(x-1)}dx$.

Posted by haifeng on 2016-04-29 04:43:34 last update 2016-04-29 04:43:34 | Answers (1) | 收藏

\[
\int_1^3 \sqrt{(3-x)(x-1)}dx
\]

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