次の曲線と2直線、およびx軸で囲まれた2つの部分の面積の和 $S$ を求めよ。 (1) $y = \frac{x-1}{x-2}$, $x = -1$, $x = \frac{3}{2}$ (2) $y = e^{2x} - 1$, $x = -1$, $x = 1$ (3) $y = \sin x$, $x = \frac{\pi}{4}$, $x = \frac{3\pi}{4}$ (4) $y = \log(x-1)$, $x = \frac{3}{2}$, $x = 4$

1. 問題の内容

次の曲線と2直線、およびx軸で囲まれた2つの部分の面積の和 $S$ を求めよ。

(1) $y = \frac{x-1}{x-2}$, $x = -1$, $x = \frac{3}{2}$

(2) $y = e^{2x} - 1$, $x = -1$, $x = 1$

(3) $y = \sin x$, $x = \frac{\pi}{4}$, $x = \frac{3\pi}{4}$

(4) $y = \log(x-1)$, $x = \frac{3}{2}$, $x = 4$

2. 解き方の手順

(1)

$y = \frac{x-1}{x-2} = \frac{x-2+1}{x-2} = 1 + \frac{1}{x-2}$

求める面積は

$S = \int_{-1}^{\frac{3}{2}} |\frac{x-1}{x-2}|dx$

$x < 1$ の時 $\frac{x-1}{x-2} > 0$, $1 < x < \frac{3}{2}$の時 $\frac{x-1}{x-2} < 0$ なので

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

$S = [x + \log|x-2|]_{-1}^{1} - [x + \log|x-2|]_{1}^{\frac{3}{2}} = (1 + \log 1) - (-1 + \log 3) - (\frac{3}{2} + \log \frac{1}{2}) + (1 + \log 1) = 1 - (-1 + \log 3) - (\frac{3}{2} - \log 2) + 1 = 2 - \log 3 - \frac{3}{2} + \log 2 + 1 = \frac{3}{2} + \log \frac{2}{3}$

(2)

$y = e^{2x} - 1$

求める面積は

$S = \int_{-1}^{1} |e^{2x} - 1| dx$

$x < 0$ の時 $e^{2x} - 1 < 0$, $x > 0$ の時 $e^{2x} - 1 > 0$ なので

$S = - \int_{-1}^{0} (e^{2x} - 1) dx + \int_{0}^{1} (e^{2x} - 1) dx = - [\frac{1}{2}e^{2x} - x]_{-1}^{0} + [\frac{1}{2}e^{2x} - x]_{0}^{1}$

$S = - (\frac{1}{2} - 0) + (\frac{1}{2}e^{-2} + 1) + (\frac{1}{2}e^{2} - 1) - (\frac{1}{2} - 0) = - \frac{1}{2} + \frac{1}{2}e^{-2} + 1 + \frac{1}{2}e^{2} - 1 - \frac{1}{2} = \frac{1}{2}e^{-2} + \frac{1}{2}e^{2} - 1 = \frac{e^2 + e^{-2}}{2} - 1 = \cosh 2 - 1$

(3)

$y = \sin x$

求める面積は

$S = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin x dx = [-\cos x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = - \cos \frac{3\pi}{4} + \cos \frac{\pi}{4} = - (-\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2} = \sqrt{2}$

(4)

$y = \log(x-1)$

求める面積は

$S = \int_{\frac{3}{2}}^{4} \log(x-1) dx$

$u = x-1$ と置換すると、$du = dx$

$S = \int_{\frac{1}{2}}^{3} \log u du = [u\log u - u]_{\frac{1}{2}}^{3} = 3\log 3 - 3 - (\frac{1}{2}\log \frac{1}{2} - \frac{1}{2}) = 3\log 3 - 3 - \frac{1}{2}(-\log 2) + \frac{1}{2} = 3\log 3 - 3 + \frac{1}{2}\log 2 + \frac{1}{2} = 3\log 3 + \frac{1}{2}\log 2 - \frac{5}{2}$

3. 最終的な答え

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

(2) $\cosh 2 - 1$

(3) $\sqrt{2}$

(4) $3\log 3 + \frac{1}{2}\log 2 - \frac{5}{2}$