以下の4つの定積分の値を求める問題です。ここで、$a$ は正の定数です。 (1) $\int_{0}^{5} (2x+6) \, dx$ (2) $\int_{1}^{2} (x^2 + \frac{1}{x^2}) \, dx$ (3) $\int_{0}^{a} \frac{1}{x^2 + a^2} \, dx$ (4) $\int_{0}^{\frac{a}{2}} \frac{1}{\sqrt{a^2 - x^2}} \, dx$

1. 問題の内容

以下の4つの定積分の値を求める問題です。ここで、$a$ は正の定数です。

(1) $\int_{0}^{5} (2x+6) \, dx$

(2) $\int_{1}^{2} (x^2 + \frac{1}{x^2}) \, dx$

(3) $\int_{0}^{a} \frac{1}{x^2 + a^2} \, dx$

(4) $\int_{0}^{\frac{a}{2}} \frac{1}{\sqrt{a^2 - x^2}} \, dx$

2. 解き方の手順

(1) $\int_{0}^{5} (2x+6) \, dx$

まず、不定積分を求めます。

$\int (2x+6) \, dx = x^2 + 6x + C$

次に、定積分を計算します。

$\int_{0}^{5} (2x+6) \, dx = [x^2 + 6x]_{0}^{5} = (5^2 + 6\cdot 5) - (0^2 + 6\cdot 0) = (25 + 30) - 0 = 55$

(2) $\int_{1}^{2} (x^2 + \frac{1}{x^2}) \, dx$

まず、不定積分を求めます。

$\int (x^2 + \frac{1}{x^2}) \, dx = \int (x^2 + x^{-2}) \, dx = \frac{1}{3}x^3 - \frac{1}{x} + C$

次に、定積分を計算します。

$\int_{1}^{2} (x^2 + \frac{1}{x^2}) \, dx = [\frac{1}{3}x^3 - \frac{1}{x}]_{1}^{2} = (\frac{1}{3}(2^3) - \frac{1}{2}) - (\frac{1}{3}(1^3) - \frac{1}{1}) = (\frac{8}{3} - \frac{1}{2}) - (\frac{1}{3} - 1) = \frac{8}{3} - \frac{1}{2} - \frac{1}{3} + 1 = \frac{7}{3} + \frac{1}{2} = \frac{14+3}{6} = \frac{17}{6}$

(3) $\int_{0}^{a} \frac{1}{x^2 + a^2} \, dx$

$\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$

$\int_{0}^{a} \frac{1}{x^2 + a^2} \, dx = [\frac{1}{a} \arctan(\frac{x}{a})]_{0}^{a} = \frac{1}{a} \arctan(\frac{a}{a}) - \frac{1}{a} \arctan(\frac{0}{a}) = \frac{1}{a} \arctan(1) - \frac{1}{a} \arctan(0) = \frac{1}{a} \cdot \frac{\pi}{4} - \frac{1}{a} \cdot 0 = \frac{\pi}{4a}$

(4) $\int_{0}^{\frac{a}{2}} \frac{1}{\sqrt{a^2 - x^2}} \, dx$

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin(\frac{x}{a}) + C$

$\int_{0}^{\frac{a}{2}} \frac{1}{\sqrt{a^2 - x^2}} \, dx = [\arcsin(\frac{x}{a})]_{0}^{\frac{a}{2}} = \arcsin(\frac{a/2}{a}) - \arcsin(\frac{0}{a}) = \arcsin(\frac{1}{2}) - \arcsin(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6}$

3. 最終的な答え

(1) 55

(2) $\frac{17}{6}$

(3) $\frac{\pi}{4a}$

(4) $\frac{\pi}{6}$