曲線の長さを求める問題

■問題

曲線 $y = \frac{1}{2} x^{2}$ $y = \frac{1}{2} x^{2}$ $(0 ≦ x ≦ 1)$ $(0 ≦ x ≦ 1)$ の長さを求めよ．

■解説動画

◇積分の動画一覧のページへ

■答

$\frac{\sqrt{2}}{2} + \frac{1}{2} \log (1 + \sqrt{2})$ $\frac{\sqrt{2}}{2} + \frac{1}{2} \log (1 + \sqrt{2})$

ヒント

曲線の長さを参考にする

■解説

図の赤の実線部分の長さを求める．

$y = \frac{1}{2} x^{2}$ $y = \frac{1}{2} x^{2}$ ， $y^{'} = x$ $y^{'} = x$

よって，曲線の長さ $L$ $L$ は公式より

$L = \int_{0}^{1} \sqrt{1 + x^{2}} d x$ $L = \int_{0}^{1} \sqrt{1 + x^{2}} d x$

となる． $x = \tan t$ $x = \tan t$ とおく置換積分にによって値を求める．

$\frac{d x}{d t} = \frac{1}{\cos^{2} t}$ $\frac{d x}{d t} = \frac{1}{\cos^{2} t}$ →　 $d x = \frac{1}{\cos^{2} t} d t$ $d x = \frac{1}{\cos^{2} t} d t$

$x$ $x$	$0 \to 1$ $0 \to 1$
$t$ $t$	$0 \to \frac{π}{4}$ $0 \to \frac{π}{4}$

よって

$= \int_{0}^{\frac{π}{4}} \sqrt{1 + \tan^{2} t} \cdot \frac{1}{\cos^{2} t} d t$ $= \int_{0}^{\frac{π}{4}} \sqrt{1 + \tan^{2} t} \cdot \frac{1}{\cos^{2} t} d t$

$= \int_{0}^{\frac{π}{4}} \sqrt{\frac{1}{\cos^{2} t}} \cdot \frac{1}{\cos^{2} t} d t$ $= \int_{0}^{\frac{π}{4}} \sqrt{\frac{1}{\cos^{2} t}} \cdot \frac{1}{\cos^{2} t} d t$

$0 < t < \frac{π}{4}$ $0 < t < \frac{π}{4}$ では， $\cos t > 0$ $\cos t > 0$ ．よって

$= \int_{0}^{\frac{π}{4}} \frac{1}{\cos t} \cdot \frac{1}{\cos^{2} t} d t$ $= \int_{0}^{\frac{π}{4}} \frac{1}{\cos t} \cdot \frac{1}{\cos^{2} t} d t$

$= \int_{0}^{\frac{π}{4}} \frac{1}{\cos^{3} t} d t$ $= \int_{0}^{\frac{π}{4}} \frac{1}{\cos^{3} t} d t$

$= \int_{0}^{\frac{π}{4}} \frac{\cos t}{\cos^{4} t} d t$ $= \int_{0}^{\frac{π}{4}} \frac{\cos t}{\cos^{4} t} d t$

$= \int_{0}^{\frac{π}{4}} \frac{\cos t}{{(1 - \sin^{2} t)}^{2}} d t$ $= \int_{0}^{\frac{π}{4}} \frac{\cos t}{{(1 - \sin^{2} t)}^{2}} d t$

$u = \sin t$ $u = \sin t$ とおく．

$\frac{d u}{d t} = \cos t$ $\frac{d u}{d t} = \cos t$ →　 $\cos t d t = d u$ $\cos t d t = d u$

$t$ $t$	$0 \to \frac{π}{4}$ $0 \to \frac{π}{4}$
$u$ $u$	$0 \to \frac{1}{\sqrt{2}}$ $0 \to \frac{1}{\sqrt{2}}$

よって

$= \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{{(1 - u^{2})}^{2}} d u$ $= \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{{(1 - u^{2})}^{2}} d u$

$= \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{{(1 - u)}^{2} {(1 + u)}^{2}} d u$ $= \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{{(1 - u)}^{2} {(1 + u)}^{2}} d u$

ここで，被積分関数を部分分数に分解する．

$\frac{1}{{(1 - u)}^{2} {(1 + u)}^{2}} = \frac{a}{{(1 - u)}^{2}} + \frac{b}{1 - u} + \frac{c}{{(1 + u)}^{2}} + \frac{d}{1 + u}$ $\frac{1}{{(1 - u)}^{2} {(1 + u)}^{2}} = \frac{a}{{(1 - u)}^{2}} + \frac{b}{1 - u} + \frac{c}{{(1 + u)}^{2}} + \frac{d}{1 + u}$

$= \frac{a + b (1 - u)}{{(1 - u)}^{2}} + \frac{c + d (1 + u)}{{(1 + u)}^{2}}$ $= \frac{a + b (1 - u)}{{(1 - u)}^{2}} + \frac{c + d (1 + u)}{{(1 + u)}^{2}}$

$= \frac{(a + b) - b u}{{(1 - u)}^{2}} + \frac{(c + d) + d u}{{(1 + u)}^{2}}$ $= \frac{(a + b) - b u}{{(1 - u)}^{2}} + \frac{(c + d) + d u}{{(1 + u)}^{2}}$

$= \frac{{(a + b) - b u} {(1 + u)}^{2} + {(c + d) + d u} {(1 - u)}^{2}}{{(1 - u)}^{2} {(1 + u)}^{2}}$ $= \frac{{(a + b) - b u} {(1 + u)}^{2} + {(c + d) + d u} {(1 - u)}^{2}}{{(1 - u)}^{2} {(1 + u)}^{2}}$

$= \frac{{(a + b) - b u} (1 + 2 u + u^{2}) + {(c + d) + d u} (1 - 2 u + u^{2})}{{(1 - u)}^{2} {(1 + u)}^{2}}$ $= \frac{{(a + b) - b u} (1 + 2 u + u^{2}) + {(c + d) + d u} (1 - 2 u + u^{2})}{{(1 - u)}^{2} {(1 + u)}^{2}}$

・分子の計算

$a + b + 2 (a + b) u + (a + b) u^{2}$ $a + b + 2 (a + b) u + (a + b) u^{2}$ $- b u - 2 b u^{2} - b u^{3}$ $- b u - 2 b u^{2} - b u^{3}$ $c + d - 2 (c + d) u + (c + d) u^{2}$ $c + d - 2 (c + d) u + (c + d) u^{2}$ $d u - 2 d u^{2} + d u^{3}$ $d u - 2 d u^{2} + d u^{3}$

$= a + b + c + d$ $= a + b + c + d$ $+ (2 a + b - 2 c - d) t$ $+ (2 a + b - 2 c - d) t$ $+ (a - b + c - d) t^{2}$ $+ (a - b + c - d) t^{2}$ $+ (- b + d) t^{3}$ $+ (- b + d) t^{3}$

ここで

$a + b + c + d = 1$ $a + b + c + d = 1$ ･･････(1)

$2 a + b - 2 c - d = 0$ $2 a + b - 2 c - d = 0$ ･･････(2)

$a - b + c - d = 0$ $a - b + c - d = 0$ ･･････(3)

$- b + d = 0$ $- b + d = 0$ ･･････(4)

である．(4)より

$d = b$ $d = b$ ･･････(5)

(5)を(2)に代入

$2 a + b - 2 c - b = 0$ $2 a + b - 2 c - b = 0$

$a - c = 0$ $a - c = 0$

$a = c$ $a = c$ ･･････(6)

(5)を(3)に代入

$a - b + c - b = 0$ $a - b + c - b = 0$

$a + c - 2 b = 0$ $a + c - 2 b = 0$

これに(6)を代入

$a + a - 2 b = 0$ $a + a - 2 b = 0$

$2 a - 2 b = 0$ $2 a - 2 b = 0$

$a = b$ $a = b$ ･･････(7)

(7)と(5)より

$d = a$ $d = a$ ･･････(8)

(6)，(7)，(8)を(1)に代入

$a + a + a + a = 1$ $a + a + a + a = 1$

$4 a = 1$ $4 a = 1$

$a = \frac{1}{4}$ $a = \frac{1}{4}$

よって

$a = b = c = d = \frac{1}{4}$ $a = b = c = d = \frac{1}{4}$

となる．したがって

$= \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{{(1 - u)}^{2} {(1 + u)}^{2}}$ $= \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{{(1 - u)}^{2} {(1 + u)}^{2}}$

$= \frac{1}{4} \int_{0}^{\frac{1}{\sqrt{2}}} {\frac{1}{{(1 - u)}^{2}} + \frac{1}{1 - u} + \frac{1}{{(1 + u)}^{2}} + \frac{1}{1 + u}} d x$ $= \frac{1}{4} \int_{0}^{\frac{1}{\sqrt{2}}} {\frac{1}{{(1 - u)}^{2}} + \frac{1}{1 - u} + \frac{1}{{(1 + u)}^{2}} + \frac{1}{1 + u}} d x$

$= \frac{1}{4} {[\frac{1}{1 - u} - \log | 1 - u | - \frac{1}{1 + u} + \log | 1 + u |]}_{0}^{\frac{1}{\sqrt{2}}}$ $= \frac{1}{4} {[\frac{1}{1 - u} - \log | 1 - u | - \frac{1}{1 + u} + \log | 1 + u |]}_{0}^{\frac{1}{\sqrt{2}}}$

$= \frac{1}{4} (\frac{1}{1 - \frac{1}{\sqrt{2}}} - \log | 1 - \frac{1}{\sqrt{2}} | - \frac{1}{1 + \frac{1}{\sqrt{2}}} + \log | 1 + \frac{1}{\sqrt{2}} | - 1 + 1)$ $= \frac{1}{4} (\frac{1}{1 - \frac{1}{\sqrt{2}}} - \log | 1 - \frac{1}{\sqrt{2}} | - \frac{1}{1 + \frac{1}{\sqrt{2}}} + \log | 1 + \frac{1}{\sqrt{2}} | - 1 + 1)$

$= \frac{1}{4} {\frac{\sqrt{2}}{\sqrt{2} - 1} - \log (\frac{\sqrt{2} - 1}{\sqrt{2}}) - \frac{\sqrt{2}}{\sqrt{2} + 1} + \log (\frac{\sqrt{2} + 1}{\sqrt{2}})}$ $= \frac{1}{4} {\frac{\sqrt{2}}{\sqrt{2} - 1} - \log (\frac{\sqrt{2} - 1}{\sqrt{2}}) - \frac{\sqrt{2}}{\sqrt{2} + 1} + \log (\frac{\sqrt{2} + 1}{\sqrt{2}})}$

$= \frac{1}{4} {\frac{\sqrt{2} (\sqrt{2} + 1) - \sqrt{2} (\sqrt{2} - 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)} + \log \frac{\sqrt{2} + 1}{\sqrt{2} - 1}}$ $= \frac{1}{4} {\frac{\sqrt{2} (\sqrt{2} + 1) - \sqrt{2} (\sqrt{2} - 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)} + \log \frac{\sqrt{2} + 1}{\sqrt{2} - 1}}$

$= \frac{1}{4} {\frac{2 + \sqrt{2} - 2 + \sqrt{2}}{2 - 1} + \log \frac{2 + 2 \sqrt{2} + 1}{2 - 1}}$ $= \frac{1}{4} {\frac{2 + \sqrt{2} - 2 + \sqrt{2}}{2 - 1} + \log \frac{2 + 2 \sqrt{2} + 1}{2 - 1}}$

$= \frac{1}{4} {2 \sqrt{2} + \log (3 + 2 \sqrt{2})}$ $= \frac{1}{4} {2 \sqrt{2} + \log (3 + 2 \sqrt{2})}$

$= \frac{\sqrt{2}}{2} + \frac{1}{4} \log (3 + 2 \sqrt{2})$ $= \frac{\sqrt{2}}{2} + \frac{1}{4} \log (3 + 2 \sqrt{2})$

$= \frac{\sqrt{2}}{2} + \frac{1}{4} \log {(\sqrt{2} + 1)}^{2}$ $= \frac{\sqrt{2}}{2} + \frac{1}{4} \log {(\sqrt{2} + 1)}^{2}$

$= \frac{\sqrt{2}}{2} + \frac{1}{2} \log (\sqrt{2} + 1)$ $= \frac{\sqrt{2}}{2} + \frac{1}{2} \log (\sqrt{2} + 1)$

■別解

$L = \int_{0}^{1} \sqrt{1 + x^{2}} d x$ $L = \int_{0}^{1} \sqrt{1 + x^{2}} d x$

部分積分法を用いて計算をする．

$= \int_{0}^{1} {(x)}^{'} \sqrt{1 + x^{2}} d x$ $= \int_{0}^{1} {(x)}^{'} \sqrt{1 + x^{2}} d x$

$= {[x \sqrt{1 + x^{2}}]}_{0}^{1} - \int_{0}^{1} x \frac{1}{2} \frac{2 x}{\sqrt{1 + x^{2}}} d x$ $= {[x \sqrt{1 + x^{2}}]}_{0}^{1} - \int_{0}^{1} x \frac{1}{2} \frac{2 x}{\sqrt{1 + x^{2}}} d x$

$= \sqrt{2} - \int_{0}^{1} \frac{x^{2}}{\sqrt{1 + x^{2}}} d x$ $= \sqrt{2} - \int_{0}^{1} \frac{x^{2}}{\sqrt{1 + x^{2}}} d x$

$= \sqrt{2} - \int_{0}^{1} \frac{1 + x^{2} - 1}{\sqrt{1 + x^{2}}} d x$ $= \sqrt{2} - \int_{0}^{1} \frac{1 + x^{2} - 1}{\sqrt{1 + x^{2}}} d x$

$= \sqrt{2} - \int_{0}^{1} \sqrt{1 + x^{2}} d x + \int_{0}^{1} \frac{1}{\sqrt{1 + x^{2}}} d x$ $= \sqrt{2} - \int_{0}^{1} \sqrt{1 + x^{2}} d x + \int_{0}^{1} \frac{1}{\sqrt{1 + x^{2}}} d x$

$= \sqrt{2} - L + {[\log | x + \sqrt{1 + x^{2}} |]}_{0}^{1}$ $= \sqrt{2} - L + {[\log | x + \sqrt{1 + x^{2}} |]}_{0}^{1}$

$= \sqrt{2} - L + \log (1 + \sqrt{2})$ $= \sqrt{2} - L + \log (1 + \sqrt{2})$

よって

$2 L = \sqrt{2} + \log (1 + \sqrt{2})$ $2 L = \sqrt{2} + \log (1 + \sqrt{2})$

$L = \frac{\sqrt{2}}{2} + \frac{1}{2} \log (1 + \sqrt{2})$ $L = \frac{\sqrt{2}}{2} + \frac{1}{2} \log (1 + \sqrt{2})$

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最終更新日：2025年2月21日

曲線の長さを求める問題

■問題

■解説動画

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■答

ヒント

■解説

■別解

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