与えられた定積分の値を求める問題です。 (1) $\int_1^4 |x^3 - 9x| dx$ (2) $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin\theta| d\theta$ (3) $\int_{-4}^{5} |\sqrt{x+4} - 2| dx$ (4) $\int_0^2 |e^x - e| dx$

1. 問題の内容

与えられた定積分の値を求める問題です。

(1) $\int_1^4 |x^3 - 9x| dx$

(2) $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin\theta| d\theta$

(3) $\int_{-4}^{5} |\sqrt{x+4} - 2| dx$

(4) $\int_0^2 |e^x - e| dx$

2. 解き方の手順

(1) $\int_1^4 |x^3 - 9x| dx$

$f(x) = x^3 - 9x = x(x^2 - 9) = x(x-3)(x+3)$

$f(x)$ の符号は $1 \le x \le 4$ で $x=3$ で変わります。

したがって、$1 \le x \le 3$ で $f(x) \le 0$ であり、$3 \le x \le 4$ で $f(x) \ge 0$ です。

$\int_1^4 |x^3 - 9x| dx = -\int_1^3 (x^3 - 9x) dx + \int_3^4 (x^3 - 9x) dx$

$= -\left[ \frac{x^4}{4} - \frac{9x^2}{2} \right]_1^3 + \left[ \frac{x^4}{4} - \frac{9x^2}{2} \right]_3^4$

$= -\left[ \frac{81}{4} - \frac{81}{2} - \left( \frac{1}{4} - \frac{9}{2} \right) \right] + \left[ \frac{256}{4} - \frac{144}{2} - \left( \frac{81}{4} - \frac{81}{2} \right) \right]$

$= -\left[ -\frac{81}{4} + \frac{17}{4} \right] + \left[ 64 - 72 - \left( -\frac{81}{4} \right) \right]$

$= \frac{64}{4} + \left[ -8 + \frac{81}{4} \right] = 16 - 8 + \frac{81}{4} = 8 + \frac{81}{4} = \frac{32+81}{4} = \frac{113}{4}$

(2) $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin\theta| d\theta$

$\sin\theta$ の符号は $\theta = 0$ で変わります。

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin\theta| d\theta = -\int_{-\frac{\pi}{4}}^0 \sin\theta d\theta + \int_0^{\frac{\pi}{4}} \sin\theta d\theta$

$= \left[ \cos\theta \right]_{-\frac{\pi}{4}}^0 - \left[ \cos\theta \right]_0^{\frac{\pi}{4}} = \left[ 1 - \frac{1}{\sqrt{2}} \right] - \left[ \frac{1}{\sqrt{2}} - 1 \right]$

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

(3) $\int_{-4}^{5} |\sqrt{x+4} - 2| dx$

$\sqrt{x+4} - 2 = 0 \Leftrightarrow \sqrt{x+4} = 2 \Leftrightarrow x+4 = 4 \Leftrightarrow x = 0$

$\int_{-4}^{5} |\sqrt{x+4} - 2| dx = -\int_{-4}^0 (\sqrt{x+4} - 2) dx + \int_0^5 (\sqrt{x+4} - 2) dx$

$= -\left[ \frac{2}{3} (x+4)^{\frac{3}{2}} - 2x \right]_{-4}^0 + \left[ \frac{2}{3} (x+4)^{\frac{3}{2}} - 2x \right]_0^5$

$= -\left[ \frac{2}{3} (4)^{\frac{3}{2}} - 0 - (0 - (-8)) \right] + \left[ \frac{2}{3} (9)^{\frac{3}{2}} - 10 - (\frac{2}{3} (4)^{\frac{3}{2}} - 0) \right]$

$= -\left[ \frac{2}{3} (8) - 8 \right] + \left[ \frac{2}{3} (27) - 10 - \frac{2}{3} (8) \right]$

$= -\left[ \frac{16}{3} - 8 \right] + \left[ 18 - 10 - \frac{16}{3} \right] = -\frac{16}{3} + 8 + 8 - \frac{16}{3} = 16 - \frac{32}{3} = \frac{48 - 32}{3} = \frac{16}{3}$

(4) $\int_0^2 |e^x - e| dx$

$e^x - e = 0 \Leftrightarrow e^x = e \Leftrightarrow x = 1$

$\int_0^2 |e^x - e| dx = -\int_0^1 (e^x - e) dx + \int_1^2 (e^x - e) dx$

$= -\left[ e^x - ex \right]_0^1 + \left[ e^x - ex \right]_1^2 = -\left[ e - e - (1 - 0)e^0 \right] + \left[ e^2 - 2e - (e - e) \right]$

$= -[0-1] + [e^2 - 2e] = 1 + e^2 - 2e = (e-1)^2$

3. 最終的な答え

(1) $\frac{113}{4}$

(2) $2 - \sqrt{2}$

(3) $\frac{16}{3}$

(4) $(e-1)^2$