测试

When $a \ne 0$, there are two solutions to $ax^2 + bx + c = 0$ and they are

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ $$ f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi $$ $$ \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} $$ $$ \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{vw} $$ $$ \overrightarrow{v} = \begin{bmatrix} v_1 \\ v_2 \\ ... \\ v_n \end{bmatrix} $$

函数极值

$f(x)$在区间$[a, b]$上连续可导， 对于$x \in (x-\xi, x+\xi) \mid \xi>0$， 存在$f(\delta) > f(x)$， 则$f(\delta)$ 是$f(x)$在区间$\lbrack a, b \rbrack$上的极大值。

洛必达法则

$$ \lim_{x \to c} \frac{u(x)}{v(x)} = \lim_{x \to c} \frac{u^{\prime}(x)}{v^{\prime}(x)} $$

定积分

$$ \int_{a}^{b}f(x)d{x} = \lim_{n \to \infty}\sum_{i=1}^{n} \Delta x \cdot f(x_i) $$ $$ \Delta x = \frac{b-a}{n} $$ $$ x_i = a + \Delta x \cdot i $$

微积分基本定理

$$ \frac{d}{dx}\int_{a}^{x}f(t)dt = f(x) $$

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