HarmonyHu 多思不如养志,多言不如守静,多才不如蓄德

微积分基础

2018-05-21
AI

导数

导数定义

\[f'(x) = \lim_{\Delta{x}->0}\frac{f(x+\Delta{x})- f(x)}{\Delta{x}} = \lim_{x->x_0}\frac{f(x)-f(x_0)}{x-x_0}\] \[f'(x)也常表示为:\frac{\alpha y}{\alpha x} 或者 \frac{\Delta y}{\Delta x} 或者 \frac{d y}{d x}\]

意义:

  1. 可以代表曲线y=f(x)在某点的切线斜率

  2. 可以反映y在x的某点上的变化率

  3. 可以表示运动曲线s=f(t)在t的某点上的速率

导数求导法则

基本运算(加减乘除)

\[[u(x) \pm v(x) ]' = u'(x) \pm v'(x)\] \[[u(x) \times v(x)]' = u'(x) \times v(x) + u(x) \times v'(x)\] \[[\frac{u(x)}{v(x)} ]' = \frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)}\]

链式法则(复合函数)

\[若y = f(u) 且u = g(x),则\frac{d y}{d x} = f'(u) \times g'(x) 或\frac{d y}{d u} \times \frac{d u}{d x}\]

初等函数的导数

\[(C)' = 0\] \[(x^n)' = nx^{n-1}\] \[(e^x)' = e^x\] \[(\sin x)' = \cos x\] \[(\cos x)' = - \sin x\] \[(a^x)' = a^x \ln a\] \[(\log{_a}x)' = \frac{1}{x \ln a} => (\ln{x})' = \frac{1}{x}\] \[f(x) = \frac{1}{1+e^{-x}} => f'(x) = f(x) \times (1-f(x))\]

求导试题

求函数导数 \(f(x) = \frac{1}{1+e^{-x}}\)

解:依据链式法则,

\[(1+e^{-x})^{-1} = (-1)\times (1+e^{-x})^{-2} \times (1+e^{-x})' = (-1)\times (1+e^{-x})^{-2} \times e^{-x} \times (-1) = \frac{1}{e^x+e^{-x}+2}\]

积分

积分定义

原函数

\[F'(x) = f(x), 称F(x)是f(x)在区间I上的原函数\]

不定积分公式

\[\int{f(x)}dx = F(x) + C, (C为常数)\]

牛顿-莱布尼茨公式

\[\int _a ^b f(x) dx = F(b) - F(a)\]

基本积分公式

\[\int x^adx = \frac{1}{a+1} x^{a+1} + C, (C是常数,a\ne -1)\] \[\int a^x dx = \frac{a^x}{\ln a} + C\] \[\int e^x dx = e^x + C\] \[\int \frac{1}{x} dx = \ln |x| + C\]

几何意义

求 \(y=x^\frac{1}{2}\) 与 \(y=x^2\) 所围图形的面积。

解:
\(A = \int_0^1(x^\frac{1}{2} - x^2) dx = [\frac{2}{3}x^\frac{3}{2}-\frac{x^3}{3}]_0^1 = \frac{1}{3}\)

积分求解方法

基本性质

\[\int kf(x) dx = k \int f(x) dx, k为常数\] \[\int [f(x)\pm g(x)]dx = \int f(x) dx \pm \int g(x) dx\] \[\int _a ^b f(x) dx = \int _a ^c f(x) dx + \int _c ^b f(x) dx\]

第一换元法

公式:
\[\int f(x)dx = \int g(u(x))u'(x)dx = \int g(u(x)) du(x) = G(u(x)) + C\]
求解试题

题1: \(f(x) = \int (ax+b)dx\) 解:
\(f(x) = \int (ax+b)dx = \int (ax+b) \times \frac{1}{a} d(ax+b) \\ 令u = ax+b,\\ 则f(x) = \frac{1}{a}\int udu = \frac{1}{2a}u^2 + C= \frac{(ax+b)^2}{2a}= \frac{a}{2}x^2 + x + C\)

**题2: ** \(f(x) = \int (3x-2)^5dx\)

解:
\(f(x) = \frac{1}{3}\int (3x-2)^5d(3x-2) +C= \frac{1}{3}\times\frac{1}{6}(3x-2)^6 +C\)

题3: \(f(x) = \int xe^{-x^2}dx\) 解:
\(\because xdx=\frac{1}{2}dx^2, \therefore f(x) = -\frac{1}{2}\int e^{-x^2}d(-x^2) = -\frac{1}{2}e^{-x^2} + C\)

第二换元法

公式:

设x=u(t), 可导且u’(t)不为0,则:
\(\int f(x)dx = \int f(u(t))u'(t)dt = F(t) + C = F(u^{-1}(x)) + C\)

求解试题
\[f(x) = \int \frac{1}{x(x-1)^{\frac{1}{2}}}dx\]

解:

\[令x=t^2+1,则\\ f(x) = \int \frac{1}{(t^2+1)t}d(t^2+1) = \int \frac{2}{t^2+1}dt = 2 \arctan t +C = 2 \arctan(x-1)^\frac{1}{2} + C\]

分步求分法

公式:
\[由[u(x)v(x)]' = u'(x)v(x) + u(x)v'(x),得u(x)v'(x) = [u(x)v(x)]'- v(x)u'(x), \\ 两边积分得\int u(x)v'(x)dx = u(x)v(x) - \int v(x)u'(x)dx\]
求解试题
\[f(x) = \int(x^2+1)e^{-x}dx\]

解:
\(f(x) = - \int (x^2+1)de^{-x} \\ = - (x^2+1)e^{-x} + \int e^{-x}d(x^2+1) \\ = - (x^2+1)e^{-x} + 2\int e^{-x}xdx \\ = - (x^2+1)e^{-x} + 2\int xd(-e^{-x}) \\ = - (x^2+1)e^{-x} - 2xe^{-x} + 2\int e^{-x}dx \\ = - (x^2+1)e^{-x} - 2xe^{-x} -2 e^{-x} +C \\ =(-x^2-2x-3)e^{-x}+C\)


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