以下の定積分を計算する問題です。 (1) $\int_{0}^{1} 2 dx$ (2) $\int_{0}^{1} x dx$ (3) $\int_{0}^{1} (x^3 + x + 1) dx$ (4) $\int_{0}^{3} 2x^2 dx$ (5) $\int_{1}^{3} 2x^2 dx$ (6) $\int_{1}^{3} (5x^2 - 1) dx$

1. 問題の内容

以下の定積分を計算する問題です。

(1) $\int_{0}^{1} 2 dx$

(2) $\int_{0}^{1} x dx$

(3) $\int_{0}^{1} (x^3 + x + 1) dx$

(4) $\int_{0}^{3} 2x^2 dx$

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

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

2. 解き方の手順

各定積分を計算します。

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

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

(3) $\int_{0}^{1} (x^3 + x + 1) dx = [\frac{1}{4}x^4 + \frac{1}{2}x^2 + x]_{0}^{1} = (\frac{1}{4}(1)^4 + \frac{1}{2}(1)^2 + (1)) - (\frac{1}{4}(0)^4 + \frac{1}{2}(0)^2 + (0)) = \frac{1}{4} + \frac{1}{2} + 1 = \frac{1+2+4}{4} = \frac{7}{4}$

(4) $\int_{0}^{3} 2x^2 dx = 2\int_{0}^{3} x^2 dx = 2[\frac{1}{3}x^3]_{0}^{3} = 2(\frac{1}{3}(3)^3 - \frac{1}{3}(0)^3) = 2(\frac{27}{3}) = 2(9) = 18$

(5) $\int_{1}^{3} 2x^2 dx = 2\int_{1}^{3} x^2 dx = 2[\frac{1}{3}x^3]_{1}^{3} = 2(\frac{1}{3}(3)^3 - \frac{1}{3}(1)^3) = 2(\frac{27}{3} - \frac{1}{3}) = 2(\frac{26}{3}) = \frac{52}{3}$

(6) $\int_{1}^{3} (5x^2 - 1) dx = [5\frac{1}{3}x^3 - x]_{1}^{3} = [\frac{5}{3}x^3 - x]_{1}^{3} = (\frac{5}{3}(3)^3 - (3)) - (\frac{5}{3}(1)^3 - (1)) = (\frac{5}{3}(27) - 3) - (\frac{5}{3} - 1) = (45 - 3) - (\frac{5}{3} - \frac{3}{3}) = 42 - \frac{2}{3} = \frac{126 - 2}{3} = \frac{124}{3}$

3. 最終的な答え

(1) 2

(2) $\frac{1}{2}$

(3) $\frac{7}{4}$

(4) 18

(5) $\frac{52}{3}$

(6) $\frac{124}{3}$