SciTech-Mathmatics-Analysis: 定积分 求解的“十大公式”

SciTech-Mathmatics-Analysis:

定积分 求解的“十大公式”

1. Newton-Leibniz formula

\(\large \begin{array}{rl} \\
\int_{a}^{b}{f'(x) dx} =& f(b) - f(a) \\
=& \underset{n \rightarrow \infty}{\lim} \overset{ n }{\underset{k=1}{\sum}} { ( f'(x_k) \cdot \Delta{x_k} ) }, 黎曼和形式\\
=& \underset{n \rightarrow \infty}{\lim} \overset{ n }{\underset{k=1}{\sum}} { \Delta{f(x_k)} },\ 无穷微分划分形 \\
\end{array}\)

2. the geometry representation of Definition of The Definite Integral

for example,

• \(\large \int_{0}^{a}{\sqrt{a^2 - x^2} dx} =\dfrac{\pi}{4} a^2\),
  
  for \(\large y =\sqrt{a^2 - x^2}\) <=> \(\large x^2 + y^2 = a^2\),
  
  a quarter of \(\large a\ circle\) to which it's radius equal to \(\large a\)
• \(\large \int_{0}^{a}{\sqrt{2 a x - x^2} dx} = \dfrac{\pi}{4} a^2\),
  
  for \(\large y =\sqrt{-(x-a)^2 + a^2}\) <=> \(\large (x-a)^2 + y^2 = a^2\),
  
  a quarter of \(\large a\ circle\) to which it's radius equal to \(\large a\) also.

3. 奇偶对称性

4. 区间对称性

\(\large \int_{-a}^{a}{f(x) dx} =\int_{0}^{a}{[f(x) + f(-x)]dx}\)

因为 在区间 \(\large x \in [-a, a]\)上, \(\large f(x)\) 与 \(\large f(-x)\) 关于 $\large x = 0 $ 轴对称.

• \(\large \int_{-a}^{a}{f(x) dx} = \int_{-a}^{a}{f(-x) dx}\)
• \(\large \int_{-a}^{0}{f(x) dx} = \int_{0}^{a}{f(-x) dx}\)
• \(\large \int_{0}^{a}{f(x) dx} = \int_{-a}^{0}{f(-x) dx}\)

\(\large \begin{array}{lrl} \\
\therefore & \int_{0}^{a}{[f(x) + f(-x)] dx} &= \int_{0}^{a}{f(x) dx}+ \int_{0}^{a}{f(-x) dx} \\
& &= \int_{-a}^{0}{f(x) dx} + \int_{0}^{a}{f(x) dx}+ \\
& &= \int_{-a}^{a}{f(x) dx} \\
& &= \int_{-a}^{a}{f(-x) dx} \\
\end{array}\)

5. 正余弦 \(\large \int_{0}^{\frac{\pi}{2}} {f(\sin{x}) dx} = \int_{0}^{\frac{\pi}{2}} {f(\cos{x}) dx}\)

由几何表示，容易直接理解.

也可：因为在区间\(\large [0, \dfrac{\pi}{2}]\)上， \(\large \int_{0}^{\frac{\pi}{2}} {f(\sin{x}) dx}\) 与 \(\large \int_{0}^{\frac{\pi}{2}} {f(\cos{x}) dx}\) 关于 $\large x = \dfrac{\pi}{4} $ 轴对称。

6.

7. $\large \int_{0}^{\pi} {f(\sin{x}) dx} = 2 \int_{0}^{\frac{\pi}{2}} {f(\sin{x}) dx} $

8. $\large \int_{0}^{\pi} {x f(\sin{x}) dx} = \frac{\pi}{2} \int_{0}^{\pi} {f(\sin{x}) dx} $

9. 区间再现公式

\(\large \int_{a}^{b}{f(x) dx} =\int_{a}^{b}{f(a +b-x) dx} =\dfrac{1}{2} \int_{a}^{b}{[f(x) + f(a +b-x)]dx}\)

因为 \(\large f(x)\) 与 \(\large f(a +b-x)\) 关于 \(\large x = \dfrac{a +b}{2}\) 轴对称；

即在区间 \(\large k \in [0, \dfrac{b-a}{2}]\) 上, \(\large f(a + k) = f(b-k)\) ;

10.