次の定積分を求めよ。 (1) $\int_1^{27} (\frac{\sqrt[3]{x}}{x^2} + 6\sqrt{x}) dx$ (2) $\int_0^1 (e^x - 12x^3) dx$ (3) $\int_0^{\frac{\pi}{3}} (\sin x + \cos x) dx$ (4) $I = \int_0^2 |x^2 - 1| dx$, ただし、$|x^2 - 1| = \begin{cases} 1 - x^2 & (0 \le x \le 1) \\ x^2 - 1 & (1 \le x \le 2) \end{cases}$ (5) $I = \int_{\frac{1}{e}}^e |\log x| dx$, ただし、$f(x) = \log x$ の原始関数 $F(x)$ は $F(x) = x \log x - x$

はい、承知いたしました。与えられた数学の問題を解いていきます。

1. 問題の内容

次の定積分を求めよ。

(1) $\int_1^{27} (\frac{\sqrt[3]{x}}{x^2} + 6\sqrt{x}) dx$

(2) $\int_0^1 (e^x - 12x^3) dx$

(3) $\int_0^{\frac{\pi}{3}} (\sin x + \cos x) dx$

(4) $I = \int_0^2 |x^2 - 1| dx$, ただし、$|x^2 - 1| = \begin{cases} 1 - x^2 & (0 \le x \le 1) \\ x^2 - 1 & (1 \le x \le 2) \end{cases}$

(5) $I = \int_{\frac{1}{e}}^e |\log x| dx$, ただし、$f(x) = \log x$ の原始関数 $F(x)$ は $F(x) = x \log x - x$

2. 解き方の手順

(1) $\int_1^{27} (\frac{\sqrt[3]{x}}{x^2} + 6\sqrt{x}) dx = \int_1^{27} (x^{-5/3} + 6x^{1/2}) dx$

$= [\frac{x^{-2/3}}{-2/3} + 6 \frac{x^{3/2}}{3/2}]_1^{27} = [-\frac{3}{2} x^{-2/3} + 4 x^{3/2}]_1^{27}$

$= [-\frac{3}{2} (27)^{-2/3} + 4 (27)^{3/2}] - [-\frac{3}{2} (1)^{-2/3} + 4 (1)^{3/2}] = [-\frac{3}{2} (\frac{1}{9}) + 4 (3\sqrt{3})^3] - [-\frac{3}{2} + 4] = [-\frac{1}{6} + 108 \sqrt{3}] - [\frac{5}{2}] = -\frac{1}{6} + 108\sqrt{3} - \frac{15}{6} = -\frac{16}{6} + 108\sqrt{3} = -\frac{8}{3} + 108\sqrt{3}$

(2) $\int_0^1 (e^x - 12x^3) dx = [e^x - 3x^4]_0^1 = (e^1 - 3(1)^4) - (e^0 - 3(0)^4) = (e - 3) - (1 - 0) = e - 4$

(3) $\int_0^{\frac{\pi}{3}} (\sin x + \cos x) dx = [-\cos x + \sin x]_0^{\frac{\pi}{3}} = (-\cos \frac{\pi}{3} + \sin \frac{\pi}{3}) - (-\cos 0 + \sin 0) = (-\frac{1}{2} + \frac{\sqrt{3}}{2}) - (-1 + 0) = -\frac{1}{2} + \frac{\sqrt{3}}{2} + 1 = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}$

(4) $I = \int_0^2 |x^2 - 1| dx = \int_0^1 (1 - x^2) dx + \int_1^2 (x^2 - 1) dx = [x - \frac{x^3}{3}]_0^1 + [\frac{x^3}{3} - x]_1^2 = (1 - \frac{1}{3}) - (0 - 0) + (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} = 2$

(5) $I = \int_{\frac{1}{e}}^e |\log x| dx$

$\log x < 0$ if $\frac{1}{e} \le x < 1$ and $\log x > 0$ if $1 < x \le e$.

$I = \int_{\frac{1}{e}}^1 -\log x dx + \int_1^e \log x dx = [-x\log x + x]_{\frac{1}{e}}^1 + [x\log x - x]_1^e = [(-1 \log 1 + 1) - (-\frac{1}{e} \log \frac{1}{e} + \frac{1}{e})] + [(e\log e - e) - (1\log 1 - 1)] = [1 - (\frac{1}{e^2} + \frac{1}{e})] + [(e-e) - (-1)] = [1 - \frac{1}{e^2} - \frac{1}{e}] + [1] = 2 - \frac{1}{e} - \frac{1}{e^2}$

3. 最終的な答え

(1) $-\frac{8}{3} + 108\sqrt{3}$

(2) $e - 4$

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

(4) $2$

(5) $2 - \frac{1}{e} - \frac{1}{e^2}$