積分法 3直線で囲まれる図形の面積、定積分の計算

学習環境

微分積分学 (ちくま学芸文庫) (吉田洋一(著)、筑摩書房)のⅣ.(積分法)、演習問題19.の解答を求めてみる。

\begin{array}{l} y = - x + 1 \\ y = x \\ y = \frac{1}{2} x \end{array}

\begin{array}{l} - x + 1 = \frac{1}{2} x \\ x = \frac{2}{3} \end{array}

\begin{array}{l} - x + 1 = x \\ x = \frac{1}{2} \end{array}

よって求める図形の面積は、

\int_{0}^{\frac{1}{2}} (x - \frac{1}{2} x) dx + \int_{\frac{1}{2}}^{\frac{2}{3}} ((- x + 1) - \frac{1}{2} x) dx

= \frac{1}{2} \int_{0}^{\frac{1}{2}} x dx + \int_{\frac{1}{2}}^{\frac{2}{3}} (1 - \frac{3}{2} x) dx

= \frac{1}{2} \cdot \frac{1}{2} \cdot {(\frac{1}{2})}^{2} + (\frac{2}{3} - \frac{1}{2}) - \frac{3}{2} \cdot \frac{1}{2} ({(\frac{2}{3})}^{2} - {(\frac{1}{2})}^{2})

= \frac{1}{16} + \frac{1}{6} - \frac{3}{4} (\frac{4}{9} - \frac{1}{4})

= \frac{1}{16} + \frac{1}{6} - \frac{3}{4} \cdot \frac{7}{36}

= \frac{1}{16} + \frac{1}{6} - \frac{7}{48}

= \frac{3 + 8 - 7}{48}

= \frac{4}{48}

= \frac{1}{12}

コード

#!/usr/bin/env python3
from unittest import TestCase, main
from sympy import Integral, Rational, solve, pprint, plot
from sympy.abc import x

print('19.')

y1 = 1 - x
y2 = x
y3 = x / 2
ys = [y1, y2, y3]
for yn in ys:
    for ym in ys:
        if yn != ym:
            pprint(solve(yn - ym, x))


class Test(TestCase):
    def test(self):
        self.assertEqual(
            Integral(x - x / 2, (x, 0, Rational(1, 2))).doit() +
            Integral(1 - 3 * x / 2, (x, Rational(1, 2), Rational(2, 3))).doit(),
            Rational(1, 12)
        )


colors = ['red', 'green', 'blue', 'brown', 'orange',
          'purple', 'pink', 'gray', 'skyblue', 'yellow']

p = plot(*ys, (x, 0, 1), legend=True, show=False)
for o, color in zip(p, colors):
    o.line_color = color
p.show()
p.save('sample19.png')


if __name__ == "__main__":
    main()

入出力結果

% ./sample19.py -v
19.
[1/2]
[2/3]
[1/2]
[0]
[2/3]
[0]
test (__main__.Test) ... ok

----------------------------------------------------------------------
Ran 1 test in 0.034s

OK
%