以下の6つの定積分を計算します。 (1) $\int_{0}^{1/2} e^{1-4x} dx$ (2) $\int_{0}^{\pi/2} \cos(x - \frac{\pi}{4}) dx$ (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) $\int_{0}^{\sqrt[3]{\pi}} x^2 \sin(x^3 + \pi) dx$

はい、承知いたしました。それでは、与えられた積分問題を解いていきます。

1. 問題の内容

以下の6つの定積分を計算します。

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

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

(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

それぞれの積分について、以下の手順で計算します。

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

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

積分範囲は、$x=0$ のとき $u=1$, $x=1/2$ のとき $u = 1-4(1/2) = -1$

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

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

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

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

$u = 4-x$ と置換すると、$x = 4-u$、$dx = -du$

積分範囲は、$x=1$ のとき $u=3$, $x=3$ のとき $u = 1$

$\int_{1}^{3} \frac{x+2}{(4-x)^3} dx = \int_{3}^{1} \frac{(4-u)+2}{u^3} (-du) = \int_{1}^{3} \frac{6-u}{u^3} du = \int_{1}^{3} (6u^{-3} - u^{-2}) du = [6 \frac{u^{-2}}{-2} - \frac{u^{-1}}{-1}]_{1}^{3} = [-3u^{-2} + u^{-1}]_{1}^{3} = (-\frac{3}{9} + \frac{1}{3}) - (-3 + 1) = (-\frac{1}{3} + \frac{1}{3}) - (-2) = 0 + 2 = 2$

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

$u = x+2$ と置換すると、$x = u-2$、$dx = du$

積分範囲は、$x=-2$ のとき $u=0$, $x=0$ のとき $u = 2$

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

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

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

積分範囲は、$x=0$ のとき $u=1$, $x=\sqrt{3}$ のとき $u = (\sqrt{3})^2 + 1 = 3+1 = 4$

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

(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 $x^2 dx = \frac{1}{3} du$

積分範囲は、$x=0$ のとき $u=\pi$, $x=\sqrt[3]{\pi}$ のとき $u = (\sqrt[3]{\pi})^3 + \pi = \pi + \pi = 2\pi$

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

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