数学 - Python - 面積、体積、長さ - 積分法の応用 – 面積 - 面積の公式(曲線や直線によって囲まれる図形の面積)

数学読本〈5〉

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
• 参考書籍
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

数学読本〈5〉微分法の応用/積分法/積分法の応用/行列と行列式(松坂 和夫(著)、岩波書店)の第20章(面積、体積、長さ - 積分法の応用)、20.1(面積)、面積の公式、問1.を取り組んでみる。

1. 1. 
      $\begin{array}{l}y=\left(x+1\right)\left(x-1\right)\\ {\displaystyle {\int }_{-1}^1\left(0-\left(x^2-1\right)\right)dx}\\ =-2{\displaystyle {\int }_0^1{x}^{2}dx}+2{\displaystyle {\int }_0^{1}1dx}\\ =-2{\left[\frac{1}{3}{x}^3\right]}_0^1+2{\left[x\right]}_0^1\\ =-\frac{2}{3}+2\\ =\frac{4}{3}\end{array}$
   2. 
      $\begin{array}{l}y=\left(1-x\right)\left(x^2+x+1\right)\\ {\displaystyle {\int }_0^1\left(1-x^3\right)dx}\\ ={\left[x-\frac{1}{4}{x}^4\right]}_0^1\\ =1-\frac{1}{4}\\ =\frac{3}{4}\end{array}$
   3. 
      $\begin{array}{l}y=x^2\left(1-x\right)\\ {\displaystyle {\int }_0^1\left(x^2-x^3\right)dx}\\ ={\left[\frac{1}{3}{x}^3-\frac{1}{4}{x}^4\right]}_0^1\\ =\frac{1}{3}-\frac{1}{4}\\ =\frac{1}{12}\end{array}$
   4. 
      $\begin{array}{l}y=-\left(x^2-2x-3\right)\\ =-\left(x-3\right)\left(x+1\right)\\ {\displaystyle {\int }_{-1}^3\left(-x^2+2x+3\right)dx}\\ ={\left[-\frac{1}{3}{x}^3+x^2+3x\right]}_{-1}^3\\ =\left(-9+9+9\right)-\left(\frac{1}{3}+1-3\right)\\ =9+\frac{5}{3}\\ =\frac{32}{3}\end{array}$
   5. 
      $\begin{array}{l}{x}^2=-x^2+3x+5\\ 2x^2-3x-5=0\\ \left(x+1\right)\left(2x-5\right)=0\\ x=-1,\frac{5}{2}\\ {\displaystyle {\int }_0^2\left(\left(-x^2+3x+5\right)-x^2\right)dx}\\ ={\displaystyle {\int }_0^2\left(-2x^2+3x+5\right)dx}\\ ={\left[-\frac{2}{3}{x}^3+\frac{3}{2}{x}^2+5x\right]}_0^2\\ =-\frac{16}{3}+6+10\\ =\frac{32}{3}\end{array}$
   6. 
      $\begin{array}{l}y=\sqrt{x}\\ x^2=\sqrt{x}\\ x^4=x\\ x\left(x^3-1\right)=0\\ x=0,1\\ {\displaystyle {\int }_0^1\left(\sqrt{x}-x^2\right)dx}\\ ={\left[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{1}{3}{x}^3\right]}_0^1\\ =\frac{2}{3}-\frac{1}{3}\\ =\frac{1}{3}\end{array}$
   7. 
      $\begin{array}{l}{y}^2=3x\\ x=\frac{y^2}{3}\\ 4y-y^2=\frac{y^2}{3}\\ 12y-3y^2=y^2\\ 4y^2-12y=0\\ 4y\left(y-3\right)=0\\ y=0,3\\ {\displaystyle {\int }_0^3\left(\left(4y-y^2\right)-\frac{y^2}{3}\right)dy}\\ ={\displaystyle {\int }_0^3\left(4y-\frac{4}{3}{y}^2\right)dy}\\ ={\left[2y^2-\frac{4}{9}{y}^3\right]}_0^3\\ =18-12\\ =6\end{array}$
   8. 
      $\begin{array}{l}{\displaystyle {\int }_0^{\frac{\pi }{4}}\cos xdx}\\ ={\left[\sin x\right]}_0^{\frac{\pi }{4}}\\ =\sin \frac{\pi }{4}-\sin 0\\ =\frac{1}{\sqrt{2}}\end{array}$
   9. 
      $\begin{array}{l}\sin x=\cos x\\ x=\frac{\pi }{4}\\ {\displaystyle {\int }_0^{\frac{\pi }{4}}\left(\cos x-\sin x\right)dx}\\ ={\left[\sin x+\cos x\right]}_0^{\frac{\pi }{4}}\\ =\left(\sin \frac{\pi }{4}+\cos \frac{\pi }{4}\right)-\left(\sin 0+\cos 0\right)\\ =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\left(0+1\right)\\ =\frac{2}{\sqrt{2}}-1\\ =\sqrt{2}-1\end{array}$
   10. 
       $\begin{array}{l}={\displaystyle {\int }_0^{\pi }\sin xdx}-{\displaystyle {\int }_{\pi }^{2\pi }\sin xdx}+{\displaystyle {\int }_{2\pi }^{3\pi }\sin xdx}\\ =2{\displaystyle {\int }_0^{\pi }\sin xdx}-{\displaystyle {\int }_{\pi }^{2\pi }\sin xdx}\\ =2{\left[-\cos x\right]}_0^{\pi }-{\left[-\cos x\right]}_{\pi }^{2\pi }\\ =2\left(-\cos \pi +\cos 0\right)+\left(\cos 2\pi -\cos \pi \right)\\ =2\left(1+1\right)+\left(1+1\right)\\ =6\end{array}$
   11. 
       $\begin{array}{l}y=\frac{1}{x}\\ y=\frac{5}{2}-x\\ \frac{5}{2}-x=\frac{1}{x}\\ \frac{5}{2}x-x^2-1=0\\ 2x^2-5x+2=0\\ x=\frac{5\pm \sqrt{25-16}}{4}\\ =\frac{5\pm \sqrt{9}}{4}\\ =\frac{5\pm 3}{4}\\ =\frac{1}{2},2\\ {\displaystyle {\int }_{\frac{1}{2}}^2\left(\left(\frac{5}{2}-x\right)-\frac{1}{x}\right)dx}\\ ={\displaystyle {\int }_{\frac{1}{2}}^2\left(\frac{5}{2}-x-\frac{1}{x}\right)dx}\\ ={\left[\frac{5}{2}x-\frac{1}{2}{x}^2-\log x\right]}_{\frac{1}{2}}^2\\ =\left(5-2-\log 2\right)-\left(\frac{5}{4}-\frac{1}{8}-\log \frac{1}{2}\right)\\ =\left(3-\log 2\right)-\left(\frac{9}{8}-\log \frac{1}{2}\right)\\ =\frac{15}{8}-\log 2+\log \frac{1}{2}\\ =\frac{15}{8}-\log 2-\log 2\\ =\frac{15}{8}-2\log 2\end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import pprint, symbols, Integral, sqrt, cos, pi, sin, Rational

print('1.')
x = symbols('x')
fs = [(0 - (x ** 2 - 1), (-1, 1)),
      (1 - x ** 3, (0, 1)),
      (x ** 2 - x ** 3, (0, 1)),
      (- x ** 2 + 2 * x + 3, (-1, 3)),
      (-x ** 2 + 3 * x + 5 - x ** 2, (0, 2)),
      (sqrt(x) - x ** 2, (0, 1)),
      ((4 * x - x ** 2) - x ** Rational(2, 3), (0, 3)),
      (cos(x), (0, pi / 4)),
      (cos(x) - sin(x), (0, pi / 4)),
      (abs(sin(x)), (0, 3 * pi)),
      (Rational(5, 2) - x - 1 / x, (Rational(1, 2), 2))]

for i, (f, (x1, x2)) in enumerate(fs, 1):
    print(f'({i})')
    I = Integral(f, (x, x1, x2))
    for g in [I, I.doit()]:
        pprint(g)
        print()
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample1.py
1.
(1)
1               
⌠               
⎮  ⎛   2    ⎞   
⎮  ⎝- x  + 1⎠ dx
⌡               
-1              

4/3


(2)
1              
⌠              
⎮ ⎛   3    ⎞   
⎮ ⎝- x  + 1⎠ dx
⌡              
0              

3/4


(3)
1               
⌠               
⎮ ⎛   3    2⎞   
⎮ ⎝- x  + x ⎠ dx
⌡               
0               

1/12


(4)
3                     
⌠                     
⎮  ⎛   2          ⎞   
⎮  ⎝- x  + 2⋅x + 3⎠ dx
⌡                     
-1                    

32/3


(5)
2                      
⌠                      
⎮ ⎛     2          ⎞   
⎮ ⎝- 2⋅x  + 3⋅x + 5⎠ dx
⌡                      
0                      

32/3


(6)
1             
⌠             
⎮ ⎛      2⎞   
⎮ ⎝√x - x ⎠ dx
⌡             
0             

1/3


(7)
3                       
⌠                       
⎮ ⎛   2/3    2      ⎞   
⎮ ⎝- x    - x  + 4⋅x⎠ dx
⌡                       
0                       

     2/3    
  9⋅3       
- ────── + 9
    5       


(8)
π          
─          
4          
⌠          
⎮ cos(x) dx
⌡          
0          

√2
──
2 


(9)
π                      
─                      
4                      
⌠                      
⎮ (-sin(x) + cos(x)) dx
⌡                      
0                      

-1 + √2


(10)
3⋅π            
 ⌠             
 ⎮  │sin(x)│ dx
 ⌡             
 0             

3⋅π            
 ⌠             
 ⎮  │sin(x)│ dx
 ⌡             
 0             


(11)
 2                 
 ⌠                 
 ⎮  ⎛     5   1⎞   
 ⎮  ⎜-x + ─ - ─⎟ dx
 ⎮  ⎝     2   x⎠   
 ⌡                 
1/2                

-2⋅log(2) + 15/8


$