与えられた6つの定積分の値を求める問題です。 (1) $\int_{0}^{\frac{1}{2}} e^{1-4x} dx$ (2) $\int_{0}^{\frac{\pi}{2}} \cos(x-\frac{\pi}{4}) dx$ (3) $\int_{1}^{3} \frac{x+2}{(4-x)^3} dx$ (4) $\int_{-2}^{0} (x-1)\sqrt{x+2} dx$ (5) $\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{x^2+1}} dx$ (6) $\int_{0}^{\sqrt[3]{\pi}} x^2 \sin(x^3+\pi) dx$

1. 問題の内容

与えられた6つの定積分の値を求める問題です。

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

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

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

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

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

(6) $\int_{0}^{\sqrt[3]{\pi}} x^2 \sin(x^3+\pi) dx$

2. 解き方の手順

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

$u = 1-4x$ と置換すると、$du = -4dx$ より $dx = -\frac{1}{4}du$。

$x=0$ のとき $u = 1$、$x=\frac{1}{2}$ のとき $u = 1-4(\frac{1}{2}) = -1$。

よって、

$\int_{0}^{\frac{1}{2}} e^{1-4x} dx = \int_{1}^{-1} e^{u} (-\frac{1}{4}) du = -\frac{1}{4} [e^u]_{1}^{-1} = -\frac{1}{4}(e^{-1} - e^1) = \frac{1}{4}(e - \frac{1}{e})$

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

$\int_{0}^{\frac{\pi}{2}} \cos(x-\frac{\pi}{4}) dx = [\sin(x-\frac{\pi}{4})]_{0}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}-\frac{\pi}{4}) - \sin(0-\frac{\pi}{4}) = \sin(\frac{\pi}{4}) - \sin(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2} - (-\frac{\sqrt{2}}{2}) = \sqrt{2}$

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

$u = 4-x$ と置換すると、$x = 4-u$, $dx = -du$。

$x=1$ のとき $u=3$, $x=3$ のとき $u=1$。

$\int_{1}^{3} \frac{x+2}{(4-x)^3} dx = \int_{3}^{1} \frac{(4-u)+2}{u^3} (-du) = \int_{1}^{3} \frac{6-u}{u^3} du = \int_{1}^{3} (6u^{-3} - u^{-2}) du = [6 \cdot \frac{u^{-2}}{-2} - \frac{u^{-1}}{-1}]_{1}^{3} = [-3u^{-2} + u^{-1}]_{1}^{3} = (-\frac{3}{9} + \frac{1}{3}) - (-3 + 1) = (-\frac{1}{3} + \frac{1}{3}) - (-2) = 2$

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

$u = x+2$ と置換すると、$x = u-2$, $dx = du$。

$x=-2$ のとき $u=0$, $x=0$ のとき $u=2$。

$\int_{-2}^{0} (x-1)\sqrt{x+2} dx = \int_{0}^{2} (u-2-1)\sqrt{u} du = \int_{0}^{2} (u-3)u^{\frac{1}{2}} du = \int_{0}^{2} (u^{\frac{3}{2}} - 3u^{\frac{1}{2}}) du = [\frac{u^{\frac{5}{2}}}{\frac{5}{2}} - 3 \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}}]_{0}^{2} = [\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}}]_{0}^{2} = (\frac{2}{5} (2^{\frac{5}{2}}) - 2 (2^{\frac{3}{2}})) - 0 = \frac{2}{5} \cdot 4\sqrt{2} - 2 \cdot 2\sqrt{2} = \frac{8\sqrt{2}}{5} - 4\sqrt{2} = \frac{8\sqrt{2} - 20\sqrt{2}}{5} = -\frac{12\sqrt{2}}{5}$

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

$u = x^2+1$ と置換すると、$du = 2xdx$ より $xdx = \frac{1}{2}du$。

$x=0$ のとき $u=1$, $x=\sqrt{3}$ のとき $u = (\sqrt{3})^2+1 = 4$。

$\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{x^2+1}} dx = \int_{1}^{4} \frac{1}{\sqrt{u}} \frac{1}{2} du = \frac{1}{2} \int_{1}^{4} u^{-\frac{1}{2}} du = \frac{1}{2} [\frac{u^{\frac{1}{2}}}{\frac{1}{2}}]_{1}^{4} = \frac{1}{2} [2\sqrt{u}]_{1}^{4} = [\sqrt{u}]_{1}^{4} = \sqrt{4} - \sqrt{1} = 2 - 1 = 1$

(6) $\int_{0}^{\sqrt[3]{\pi}} x^2 \sin(x^3+\pi) dx$

$u = x^3 + \pi$ と置換すると、$du = 3x^2 dx$ より $x^2 dx = \frac{1}{3} du$。

$x=0$ のとき $u = \pi$, $x = \sqrt[3]{\pi}$ のとき $u = (\sqrt[3]{\pi})^3 + \pi = \pi + \pi = 2\pi$。

$\int_{0}^{\sqrt[3]{\pi}} x^2 \sin(x^3+\pi) dx = \int_{\pi}^{2\pi} \sin(u) \frac{1}{3} du = \frac{1}{3} \int_{\pi}^{2\pi} \sin(u) du = \frac{1}{3} [-\cos(u)]_{\pi}^{2\pi} = \frac{1}{3} (-\cos(2\pi) - (-\cos(\pi))) = \frac{1}{3} (-1 - (-(-1))) = \frac{1}{3}(-1-1) = -\frac{2}{3}$

3. 最終的な答え

(1) $\frac{1}{4}(e - \frac{1}{e})$

(2) $\sqrt{2}$

(3) $2$

(4) $-\frac{12\sqrt{2}}{5}$

(5) $1$

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