与えられた5つの定積分の値を計算する問題です。 (1) $\int_{0}^{1} (x^2 - 3x + 1) dx$ (2) $\int_{0}^{\frac{\pi}{2}} \sin 2x dx$ (3) $\int_{0}^{1} e^{2x} dx$ (4) $\int_{0}^{3} \sqrt{x+1} dx$ (5) $\int_{2}^{4} \frac{1}{x-1} dx$

1. 問題の内容

与えられた5つの定積分の値を計算する問題です。

(1) $\int_{0}^{1} (x^2 - 3x + 1) dx$

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

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

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

(5) $\int_{2}^{4} \frac{1}{x-1} dx$

2. 解き方の手順

(1)

$\int_{0}^{1} (x^2 - 3x + 1) dx = [\frac{1}{3}x^3 - \frac{3}{2}x^2 + x]_{0}^{1} = (\frac{1}{3} - \frac{3}{2} + 1) - (0) = \frac{2 - 9 + 6}{6} = -\frac{1}{6}$

(2)

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

(3)

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

(4)

$\int_{0}^{3} \sqrt{x+1} dx = \int_{0}^{3} (x+1)^{\frac{1}{2}} dx = [\frac{2}{3}(x+1)^{\frac{3}{2}}]_{0}^{3} = \frac{2}{3}(4^{\frac{3}{2}}) - \frac{2}{3}(1^{\frac{3}{2}}) = \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}$

(5)

$\int_{2}^{4} \frac{1}{x-1} dx = [\ln|x-1|]_{2}^{4} = \ln|4-1| - \ln|2-1| = \ln 3 - \ln 1 = \ln 3 - 0 = \ln 3$

3. 最終的な答え

(1) $-\frac{1}{6}$

(2) $1$

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

(4) $\frac{14}{3}$

(5) $\ln 3$