MIT_单变量微积分_21

版权声明: https://blog.csdn.net/qq_38386316/article/details/88648375

1.功、平均值、概率

Ex:平均值:
y 1 + y 2 + . . . + y n n 1 b a z b f ( x ) d x \frac{y_1+y_2+...+y_n}{n} \to \frac{1}{b-a}\int_z^bf(x)dx

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a = x 0 < x 1 < x 2 < . . . < x n = b y 1 = f ( x 1 ) , y 2 = f ( x 2 ) , y 3 = f ( x 3 ) : ( y 1 + . . . . + y n ) Δ x b a a b f ( x ) d x b a ( Δ x ) Δ x b a = 1 n a=x_0<x_1<x_2<...<x_n=b\\ y_1=f(x_1),y_2=f(x_2),y_3=f(x_3)\\ 黎曼和:\\ \frac{(y_1+....+y_n)\Delta x}{b-a} \to \frac{\int_a^bf(x)dx}{b-a} (\Delta x \to \infty)\\ \frac{\Delta x}{b-a}=\frac{1}{n}
Ex: f ( x ) = C f(x)=C 求平均值。
1 b a a b C d x = C \frac{1}{b-a}\int_a^bCdx=C

Ex:点在单位半圆上的平均高度
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1 1 ( 1 ) 1 1 1 x 2 d x = 1 2 n 2 = π 4 \frac{1}{1-(-1)}*\int_{-1}^1\sqrt{1-x^2}dx=\frac{1}{2}*\frac{n}{2}=\frac{\pi}{4}
Ex:上图弧长的平均值 ( 0 θ π ) (0 \leq \theta \leq \pi)
1 π 0 θ s i n θ d θ = 1 π c o s θ 0 π = 1 π ( 2 ) = 2 π \frac{1}{\pi}\int_0^{\theta}sin \theta d \theta =-\frac{1}{\pi}*cos \theta |_0^{\pi}=\frac{-1}{\pi}(-2)=\frac{2}{\pi}

加权平均值:
a b f ( x ) w ( x ) d x a b w ( x ) d x \frac{\int_a^bf(x)w(x)dx}{\int_a^bw(x)dx}

  • 解释1: A V C ( C ) = C AVC(C)=C
    a b C w ( x ) d x a b w ( x ) d x = C a b w ( x ) d x a b w ( x ) d x = C \frac{\int_a^bCw(x)dx}{\int_a^bw(x)dx}=\frac{C\int_a^bw(x)dx}{\int_a^bw(x)dx}=C
  • 解释2:股票例子
    10 w 1 + 20 w 2 + 30 w 3 w 1 + w 2 + w 3 \frac{10w_1+20w_2+30w_3}{w_1+w_2+w_3}
    Ex:坩埚例子:
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    初始: T = 0 T=0
    最终: T = 100 30 y T=100-30y
    能量= 体积 * 温度
    0 1 T π x 2 d y = 0 1 ( 100 30 y ) π y d y = 0 1 100 π y 30 π y 2 d y = 50 π y 2 10 π y 3 0 1 = 40 π \int_0^1T\pi x^2dy\\ =\int_0^1(100-30y)\pi ydy\\ =\int_0^1100\pi y-30\pi y^2dy\\ =50\pi y^2-10\pi y^3|_0^1\\ =40\pi

最后的平均温度:
0 1 T π y d y 0 1 π y d y = 40 π π 2 = 8 0 0 \frac{\int_0^1T\pi y dy}{\int_0^1\pi y dy}=\frac{40 \pi}{\frac{\pi}{2}}=80^0
平常的平均温度:
T m a x + T m i n 2 = 100 + 70 2 = 8 5 0 \frac{T_{max}+T_{min}}{2}=\frac{100+70}{2}=85^0

2. 概率

Ex: 0 < y < 1 x 2 0<y<1-x^2 ,使得 y > 1 2 y>\frac{1}{2}的概率
1 2 1 ( 1 x 2 ) d x 1 1 ( 1 x 2 ) d x = P ( x > 1 2 ) \frac{\int_\frac{1}{2}^1(1-x^2)dx}{\int_{-1}^{1}(1-x^2)dx}=P(x>\frac{1}{2})
求概率的通常的公式:
a x 1 x 2 b , P ( x 1 < x < x 2 ) = x 1 x 2 w ( x ) d x a b w ( x ) d x = P a r t T o t a l a \leq x_1 \leq x_2 \leq b,P(x_1<x<x_2)=\frac{\int_{x_1}^{x_2}w(x)dx}{\int_a^bw(x)dx}=\frac{Part}{Total}
Ex:靶子问题: f = C e r 2 f=Ce^{-r^2} (模型)

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P A R T = r 1 r 2 2 π r e r 2 d r = π e r 2 r 1 r 2 = π ( e r 1 2 e r 2 2 ) P A R T = C π ( e r 1 2 e r 2 2 ) W h o l e = C π ( e 0 2 e 2 ) = C π P A R T W H O L E = e r 1 2 e r 2 2 PART=\int_{r_1}^{r_2}2\pi re^{-r^2}dr\\ =-\pi e^{-r^2}|_{r_1}^{r_2}\\ =\pi(e^{-r_1^2}-e^{-r_2^2})\\ PART=C\pi(e^{-r_1^2}-e^{-r_2^2})\\ Whole=C\pi(e^{0^2}-e^{-\infty^2})\\ =C\pi\\ \frac{PART}{WHOLE}=e^{r_1^2}-e^{r_2^2}
Ex:假设在靶子旁边站着一个人。求小人被射中的概率。 2 12 P ( 2 a < r < 3 a ) = ? \frac{2}{12}*P(2a<r<3a)=?
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P ( 0 < r < a ) = 1 2 e 0 2 e a 2 = 1 2 e a 2 = 1 2 P ( 2 a < r < 3 a ) = e ( 2 a ) 2 e ( 3 a ) 2 = e ( a 2 ) 4 e ( a 2 ) 9 = ( 1 2 ) 4 1 2 9 1 16 P(0<r<a)=\frac{1}{2}\\ e^{-0^2}-e^{-a^2}=\frac{1}{2}\\ e^{-a^2}=\frac{1}{2}\\ P(2a<r<3a)=e^{-(2a)^2}-e^{-(3a)^2}\\ =e^{(-a^2)^4}-e^{(-a^2)^9}\\ =(\frac{1}{2})^4-\frac{1}{2}^9\\ \approx \frac{1}{16}

2 12 P ( 2 a < r < 3 a ) = 1 32 \frac{2}{12}*P(2a<r<3a)=\frac{1}{32} .
权重为: w ( r ) = 2 π c t e 2 w(r)=2\pi c t e^{-2}

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转载自blog.csdn.net/qq_38386316/article/details/88648375