1.已知标准正态分布的概率密度为,
.根据概率密度的性质有:
=1.用数值积分法验证这一事实(精度为
)。
clc;clear;%梯形数值积分
t=-5:0.01:5;jifen=0;
f=inline('1/sqrt(2*pi)*exp(-x^2/2)');
for i=1:1000
jifen=jifen+(f(t(i))+f(t(i+1)))/2*0.01;
end
jifen
jifen =
0.999999426572967
2.用多种数值积分方法计算定积分,与精确值进行比较,观察不同方法对应的误差。
clc;clear;
f=@(x)1/(1-sin(x));
f1=sym(f);
%标准
right=int(f1,0,pi/4);
disp(double(right));
%梯形数值积分
t=0:0.001:pi/4;jifen=0;
for i=1:785
jifen=jifen+(f(t(i))+f(t(i+1)))/2*0.001;
end
disp(jifen);
%泰勒
jifen2=int(taylor(f1,'order',20),0,pi/4);
jifen2=double(jifen2)
%抛物线数值积分
jifen3=0;
for i=1:785
jifen3=jifen3+(f(t(i))+4*f((t(i)+t(i+1))/2)+f(t(i)))*0.001/6;
end
jifen3
标准:1.414213562373095;
梯形数值积分:1.412855403046629;
taylor=1.414212348117243;
抛物线数值积分=1.412452978182123;
3.图8.2所示是某平面区域的示意图.
在选取自西向东的方向为x轴正方向,自南向北的方向为y轴正向的情况下,测得该平面区域边界上的一些数据如表8.5所示:
表8.5 平面区域坐标数据
x |
2.0 |
8.6 |
12.0 |
21.8 |
35.4 |
42.9 |
45.6 |
61.6 |
69.8 |
87.2 |
y1 |
14.5 |
21.6 |
22.4 |
25.8 |
20.6 |
23.8 |
36.8 |
28.8 |
30.4 |
38.7 |
y2 |
94.5 |
76.8 |
91.5 |
118.6 |
124.2 |
140.6 |
172.8 |
206.6 |
175.5 |
198.6 |
x |
120.5 |
146.8 |
168.5 |
189.8 |
220.4 |
255.0 |
286.6 |
302.5 |
350.1 |
|
y1 |
52.8 |
98.6 |
55.5 |
69.4 |
23.4 |
51.9 |
46.8 |
52.8 |
66.6 |
|
y2 |
145.6 |
186.8 |
352.4 |
423.8 |
386.7 |
415.4 |
358.2 |
245.2 |
198.9 |
clc;clear;
x=[2.0,8.6,12.0,21.8,35.4,42.9,45.6,61.1,69.8,87.2,120.5,146.8,168.5,189.8,220.4,255.0,286.6,302.5,350.1];
y1=[14.5,21.6,22.4,25.8,20.6,23.8,36.8,28.8,30.4,38.7,52.8,98.6,55.5,69.4,23.4,51.9,46.8,52.8,66.6];
y2=[94.5,76.8,91.5,118.6,124.2,140.6,172.8,206.6,175.5,198.6,145.6,186.8,352.4,423.8,386.7,415.4,358.2,245.2,189.9];
plot(x,y1,x,y2);
这个题有点奇怪,为何提上说是最西端和最东段而图画出的却是非封闭图形?是我理解错了?大家觉得呢?还请高人指点。
本文由作者自创,由于时间原因,难免出现些许错误,还请大家多多指正。创作不易,望多多鼓励。