次の定積分を求めます。 (1) $\int_{1}^{e} \frac{dx}{x}$ (2) $\int_{1}^{2} \frac{dy}{y^3}$ (3) $\int_{0}^{\frac{\pi}{4}} \frac{d\theta}{\cos^2\theta}$ (4) $\int_{0}^{1} (e^x - e^{-x})dx$ (5) $\int_{1}^{2} \frac{dx}{x(x-3)}$ (6) $\int_{0}^{\pi} \cos x \sin 4x dx$ (7) $\int_{0}^{\frac{\pi}{4}} \sin^2 x dx$ (8) $\int_{0}^{2\pi} |\sin x| dx$ (9) $\int_{0}^{4} \sqrt{|x-1|} dx$ (10) $\int_{0}^{1} |e^x - 2| dx$

1. 問題の内容

次の定積分を求めます。

(1) $\int_{1}^{e} \frac{dx}{x}$

(2) $\int_{1}^{2} \frac{dy}{y^3}$

(3) $\int_{0}^{\frac{\pi}{4}} \frac{d\theta}{\cos^2\theta}$

(4) $\int_{0}^{1} (e^x - e^{-x})dx$

(5) $\int_{1}^{2} \frac{dx}{x(x-3)}$

(6) $\int_{0}^{\pi} \cos x \sin 4x dx$

(7) $\int_{0}^{\frac{\pi}{4}} \sin^2 x dx$

(8) $\int_{0}^{2\pi} |\sin x| dx$

(9) $\int_{0}^{4} \sqrt{|x-1|} dx$

(10) $\int_{0}^{1} |e^x - 2| dx$

2. 解き方の手順

(1)

$\int_{1}^{e} \frac{dx}{x} = [\ln|x|]_{1}^{e} = \ln e - \ln 1 = 1 - 0 = 1$

(2)

$\int_{1}^{2} \frac{dy}{y^3} = \int_{1}^{2} y^{-3} dy = [\frac{y^{-2}}{-2}]_{1}^{2} = [-\frac{1}{2y^2}]_{1}^{2} = -\frac{1}{2 \cdot 2^2} - (-\frac{1}{2 \cdot 1^2}) = -\frac{1}{8} + \frac{1}{2} = \frac{-1 + 4}{8} = \frac{3}{8}$

(3)

$\int_{0}^{\frac{\pi}{4}} \frac{d\theta}{\cos^2\theta} = \int_{0}^{\frac{\pi}{4}} \sec^2\theta d\theta = [\tan\theta]_{0}^{\frac{\pi}{4}} = \tan\frac{\pi}{4} - \tan 0 = 1 - 0 = 1$

(4)

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

(5)

$\frac{1}{x(x-3)} = \frac{A}{x} + \frac{B}{x-3}$

$1 = A(x-3) + Bx$

$1 = (A+B)x - 3A$

$A+B = 0$, $-3A = 1$

$A = -\frac{1}{3}$, $B = \frac{1}{3}$

$\int_{1}^{2} \frac{dx}{x(x-3)} = \int_{1}^{2} (-\frac{1}{3x} + \frac{1}{3(x-3)})dx = [-\frac{1}{3} \ln|x| + \frac{1}{3} \ln|x-3|]_{1}^{2} = [\frac{1}{3} \ln|\frac{x-3}{x}|]_{1}^{2} = \frac{1}{3} \ln|\frac{2-3}{2}| - \frac{1}{3} \ln|\frac{1-3}{1}| = \frac{1}{3} \ln\frac{1}{2} - \frac{1}{3} \ln 2 = \frac{1}{3} (\ln\frac{1}{2} - \ln 2) = \frac{1}{3} (\ln 1 - \ln 2 - \ln 2) = \frac{1}{3} (-2 \ln 2) = -\frac{2}{3} \ln 2$

(6)

$\int_{0}^{\pi} \cos x \sin 4x dx = \int_{0}^{\pi} \cos x (4 \sin x \cos x - 4 \sin^3 x) dx$

$= \int_{0}^{\pi} \cos x (4 \sin x \cos x - 4 \sin x (1 - \cos^2 x)) dx = \int_{0}^{\pi} (4\sin x \cos^2 x - 4 \sin x + 4 \sin x \cos^2 x) dx = \int_{0}^{\pi} (8 \sin x \cos^2 x - 4 \sin x \cos^4 x) dx = [-\frac{8}{3} \cos^3 x - \frac{4}{5} \cos^5 x]_{0}^{\pi} = (-\frac{8}{3}(-1)^3 - \frac{4}{5}(-1)^5) - (-\frac{8}{3}(1)^3 - \frac{4}{5}(1)^5) = (\frac{8}{3} + \frac{4}{5}) - (-\frac{8}{3} - \frac{4}{5}) = 2 (\frac{8}{3} + \frac{4}{5}) = 2 (\frac{40+12}{15}) = \frac{2 \cdot 52}{15} = \frac{104}{15}$

$\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x (2 \cos^2 x -1)$

$\int_{0}^{\pi} \frac{1}{2} [\sin(4x+x) + \sin (4x-x)]dx= \frac{1}{2} [\sin 5x - \sin 3x] = \frac{1}{2} [\frac{-\cos5x}{5} - \frac{-cos3x}{3}] = \frac{1}{2}[(\frac{-\cos5\pi}{5}+\frac{\cos3\pi}{3}) - (\frac{-\cos0}{5} + \frac{\cos0}{3})] = \frac{1}{2} [(\frac{-(-1)}{5}+\frac{-1}{3}) - (\frac{-1}{5} + \frac{1}{3})] = \frac{1}{2} [(\frac{1}{5} - \frac{1}{3}) - (\frac{-1}{5} + \frac{1}{3})] = \frac{1}{2}[(\frac{3-5}{15})-(\frac{-3+5}{15})] = \frac{1}{2} (\frac{-2}{15} - \frac{2}{15}) = \frac{-2}{15}$

(7)

$\int_{0}^{\frac{\pi}{4}} \sin^2 x dx = \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos 2x}{2} dx = [\frac{x}{2} - \frac{\sin 2x}{4}]_{0}^{\frac{\pi}{4}} = (\frac{\pi}{8} - \frac{\sin \frac{\pi}{2}}{4}) - (0 - 0) = \frac{\pi}{8} - \frac{1}{4}$

(8)

$\int_{0}^{2\pi} |\sin x| dx = \int_{0}^{\pi} \sin x dx + \int_{\pi}^{2\pi} (-\sin x) dx = [-\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{2\pi} = (-\cos\pi - (-\cos 0)) + (\cos 2\pi - \cos \pi) = (-(-1) - (-1)) + (1 - (-1)) = (1+1) + (1+1) = 2+2 = 4$

(9)

$\int_{0}^{4} \sqrt{|x-1|} dx = \int_{0}^{1} \sqrt{1-x} dx + \int_{1}^{4} \sqrt{x-1} dx = [-\frac{2}{3} (1-x)^{\frac{3}{2}}]_{0}^{1} + [\frac{2}{3} (x-1)^{\frac{3}{2}}]_{1}^{4} = (0 - (-\frac{2}{3} (1)^{\frac{3}{2}})) + (\frac{2}{3} (3)^{\frac{3}{2}} - 0) = \frac{2}{3} + \frac{2}{3} 3\sqrt{3} = \frac{2}{3} + 2\sqrt{3} = \frac{2+6\sqrt{3}}{3}$

(10)

$|e^x - 2| = e^x - 2$ when $e^x \ge 2$ or $x \ge \ln 2$

$|e^x - 2| = 2 - e^x$ when $e^x < 2$ or $x < \ln 2$

$\int_{0}^{1} |e^x - 2| dx = \int_{0}^{\ln 2} (2-e^x) dx + \int_{\ln 2}^{1} (e^x-2) dx = [2x - e^x]_{0}^{\ln 2} + [e^x - 2x]_{\ln 2}^{1} = (2 \ln 2 - e^{\ln 2}) - (0 - e^0) + (e^1 - 2 \cdot 1) - (e^{\ln 2} - 2 \ln 2) = (2 \ln 2 - 2) + 1 + (e-2) - (2 - 2 \ln 2) = 2 \ln 2 - 2 + 1 + e - 2 - 2 + 2 \ln 2 = 4 \ln 2 + e - 5$

3. 最終的な答え

(1) 1

(2) $\frac{3}{8}$

(3) 1

(4) $e + \frac{1}{e} - 2$

(5) $-\frac{2}{3} \ln 2$

(6) $-\frac{2}{15}$

(7) $\frac{\pi}{8} - \frac{1}{4}$

(8) 4

(9) $\frac{2+6\sqrt{3}}{3}$

(10) $4 \ln 2 + e - 5$