与えられた6つの定積分を計算する問題です。 (1) $\int_{1}^{2} \sqrt{x+1} \, dx$ (2) $\int_{0}^{1} (2x+1)^3 \, dx$ (3) $\int_{-1}^{1} (e^t - e^{-t}) \, dt$ (4) $\int_{0}^{\pi} \sin 2x \, dx$ (5) $\int_{0}^{2\pi} \cos^2 x \, dx$ (6) $\int_{0}^{\frac{\pi}{2}} \sin 4\theta \cos 2\theta \, d\theta$

1. 問題の内容

与えられた6つの定積分を計算する問題です。

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

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

(3) $\int_{-1}^{1} (e^t - e^{-t}) \, dt$

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

(5) $\int_{0}^{2\pi} \cos^2 x \, dx$

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

2. 解き方の手順

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

$u = x+1$ と置換すると、$du = dx$。積分範囲は$x=1$のとき$u=2$、$x=2$のとき$u=3$。

$\int_{2}^{3} \sqrt{u} \, du = \int_{2}^{3} u^{1/2} \, du = \left[ \frac{2}{3} u^{3/2} \right]_{2}^{3} = \frac{2}{3}(3^{3/2} - 2^{3/2}) = \frac{2}{3}(3\sqrt{3} - 2\sqrt{2})$

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

$u = 2x+1$ と置換すると、$du = 2dx$ より $dx = \frac{1}{2} du$。積分範囲は$x=0$のとき$u=1$、$x=1$のとき$u=3$。

$\int_{1}^{3} u^3 \frac{1}{2} \, du = \frac{1}{2} \int_{1}^{3} u^3 \, du = \frac{1}{2} \left[ \frac{1}{4} u^4 \right]_{1}^{3} = \frac{1}{8}(3^4 - 1^4) = \frac{1}{8}(81-1) = \frac{80}{8} = 10$

(3) $\int_{-1}^{1} (e^t - e^{-t}) \, dt$

$\int_{-1}^{1} (e^t - e^{-t}) \, dt = \left[ e^t + e^{-t} \right]_{-1}^{1} = (e^1 + e^{-1}) - (e^{-1} + e^1) = 0$

または、被積分関数が奇関数であることから、積分範囲が$-1$から$1$なので、積分値は$0$。

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

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

(5) $\int_{0}^{2\pi} \cos^2 x \, dx$

$\cos^2 x = \frac{1 + \cos 2x}{2}$より、

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

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

積和の公式$\sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B))$を用いると、

$\sin 4\theta \cos 2\theta = \frac{1}{2} (\sin 6\theta + \sin 2\theta)$

$\int_{0}^{\frac{\pi}{2}} \sin 4\theta \cos 2\theta \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} (\sin 6\theta + \sin 2\theta) \, d\theta = \frac{1}{2} \left[ -\frac{1}{6} \cos 6\theta - \frac{1}{2} \cos 2\theta \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left[ \left(-\frac{1}{6}\cos 3\pi - \frac{1}{2} \cos \pi \right) - \left(-\frac{1}{6}\cos 0 - \frac{1}{2}\cos 0 \right) \right] = \frac{1}{2} \left[ \left(-\frac{1}{6}(-1) - \frac{1}{2}(-1) \right) - \left(-\frac{1}{6} - \frac{1}{2} \right) \right] = \frac{1}{2} \left[ \frac{1}{6} + \frac{1}{2} + \frac{1}{6} + \frac{1}{2} \right] = \frac{1}{2} \left[ \frac{2}{6} + 1 \right] = \frac{1}{2} \left[ \frac{1}{3} + 1 \right] = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$

3. 最終的な答え

(1) $\frac{2}{3}(3\sqrt{3} - 2\sqrt{2})$

(2) $10$

(3) $0$

(4) $0$

(5) $\pi$

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