以下の4つの定積分を計算します。 (1) $\int_{-1}^{1} 5e^{x} dx$ (2) $\int_{1}^{2} (3 + \frac{2}{x})^{2} dx = \int_{1}^{2} (9 + 12x^{-1} + 4x^{-2}) dx$ (3) $\int_{\frac{\pi}{2}}^{\pi} \sin x dx$ (4) $\int_{0}^{\frac{\pi}{6}} \frac{2}{\cos^{2} x} dx$

1. 問題の内容

以下の4つの定積分を計算します。

(1) $\int_{-1}^{1} 5e^{x} dx$

(2) $\int_{1}^{2} (3 + \frac{2}{x})^{2} dx = \int_{1}^{2} (9 + 12x^{-1} + 4x^{-2}) dx$

(3) $\int_{\frac{\pi}{2}}^{\pi} \sin x dx$

(4) $\int_{0}^{\frac{\pi}{6}} \frac{2}{\cos^{2} x} dx$

2. 解き方の手順

(1)

$\int_{-1}^{1} 5e^{x} dx = [5e^{x}]_{-1}^{1} = 5e^{1} - 5e^{-1} = 5e - \frac{5}{e}$

(2)

$\int_{1}^{2} (9 + 12x^{-1} + 4x^{-2}) dx = [9x + 12\ln|x| + 4\cdot \frac{x^{-1}}{-1}]_{1}^{2} = [9x + 12\ln|x| - \frac{4}{x}]_{1}^{2} = (9(2) + 12\ln 2 - \frac{4}{2}) - (9(1) + 12\ln 1 - \frac{4}{1}) = 18 + 12\ln 2 - 2 - 9 - 0 + 4 = 11 + 12\ln 2$

(3)

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

(4)

$\int_{0}^{\frac{\pi}{6}} \frac{2}{\cos^{2} x} dx = \int_{0}^{\frac{\pi}{6}} 2\sec^{2} x dx = [2\tan x]_{0}^{\frac{\pi}{6}} = 2\tan \frac{\pi}{6} - 2\tan 0 = 2\cdot \frac{1}{\sqrt{3}} - 2\cdot 0 = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

3. 最終的な答え

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

(2) $11 + 12\ln 2$

(3) $1$

(4) $\frac{2\sqrt{3}}{3}$