以下の4つの定積分を計算する問題です。 (1) $\int_1^2 (6x^5 + 5x^4 - \frac{1}{x^2}) dx$ (2) $\int_1^4 (\sqrt{\frac{4}{x^3}} + \frac{1}{\sqrt{2x}} + (3x)^2) dx$ (3) $\int_{-2}^{-1} (x-1)(\frac{1}{x^3} + 1) dx$ (4) $\int_2^3 \frac{x^2 + \sqrt{x^3} - \sqrt{x}}{x} dx$

1. 問題の内容

以下の4つの定積分を計算する問題です。

(1) $\int_1^2 (6x^5 + 5x^4 - \frac{1}{x^2}) dx$

(2) $\int_1^4 (\sqrt{\frac{4}{x^3}} + \frac{1}{\sqrt{2x}} + (3x)^2) dx$

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

(4) $\int_2^3 \frac{x^2 + \sqrt{x^3} - \sqrt{x}}{x} dx$

2. 解き方の手順

(1)

まず、$6x^5 + 5x^4 - \frac{1}{x^2} = 6x^5 + 5x^4 - x^{-2}$ の不定積分を計算します。

$\int (6x^5 + 5x^4 - x^{-2}) dx = x^6 + x^5 + x^{-1} + C = x^6 + x^5 + \frac{1}{x} + C$

したがって、定積分は

$\int_1^2 (6x^5 + 5x^4 - \frac{1}{x^2}) dx = [x^6 + x^5 + \frac{1}{x}]_1^2 = (2^6 + 2^5 + \frac{1}{2}) - (1^6 + 1^5 + \frac{1}{1}) = (64 + 32 + \frac{1}{2}) - (1 + 1 + 1) = 96 + \frac{1}{2} - 3 = 93 + \frac{1}{2} = \frac{187}{2}$

(2)

まず、$\sqrt{\frac{4}{x^3}} + \frac{1}{\sqrt{2x}} + (3x)^2 = 2x^{-3/2} + \frac{1}{\sqrt{2}}x^{-1/2} + 9x^2$ の不定積分を計算します。

$\int (2x^{-3/2} + \frac{1}{\sqrt{2}}x^{-1/2} + 9x^2) dx = 2 \cdot \frac{x^{-1/2}}{-1/2} + \frac{1}{\sqrt{2}} \cdot \frac{x^{1/2}}{1/2} + 9 \cdot \frac{x^3}{3} + C = -4x^{-1/2} + \frac{2}{\sqrt{2}}x^{1/2} + 3x^3 + C = -\frac{4}{\sqrt{x}} + \sqrt{2x} + 3x^3 + C$

したがって、定積分は

$\int_1^4 (\sqrt{\frac{4}{x^3}} + \frac{1}{\sqrt{2x}} + (3x)^2) dx = [-\frac{4}{\sqrt{x}} + \sqrt{2x} + 3x^3]_1^4 = (-\frac{4}{\sqrt{4}} + \sqrt{2 \cdot 4} + 3 \cdot 4^3) - (-\frac{4}{\sqrt{1}} + \sqrt{2 \cdot 1} + 3 \cdot 1^3) = (-2 + \sqrt{8} + 3 \cdot 64) - (-4 + \sqrt{2} + 3) = (-2 + 2\sqrt{2} + 192) - (-1 + \sqrt{2}) = 190 + 2\sqrt{2} + 1 - \sqrt{2} = 191 + \sqrt{2}$

(3)

まず、$(x-1)(\frac{1}{x^3} + 1) = \frac{x}{x^3} + x - \frac{1}{x^3} - 1 = x^{-2} + x - x^{-3} - 1$ の不定積分を計算します。

$\int (x^{-2} + x - x^{-3} - 1) dx = \frac{x^{-1}}{-1} + \frac{x^2}{2} - \frac{x^{-2}}{-2} - x + C = -\frac{1}{x} + \frac{x^2}{2} + \frac{1}{2x^2} - x + C$

したがって、定積分は

$\int_{-2}^{-1} (x-1)(\frac{1}{x^3} + 1) dx = [-\frac{1}{x} + \frac{x^2}{2} + \frac{1}{2x^2} - x]_{-2}^{-1} = (-\frac{1}{-1} + \frac{(-1)^2}{2} + \frac{1}{2(-1)^2} - (-1)) - (-\frac{1}{-2} + \frac{(-2)^2}{2} + \frac{1}{2(-2)^2} - (-2)) = (1 + \frac{1}{2} + \frac{1}{2} + 1) - (\frac{1}{2} + 2 + \frac{1}{8} + 2) = 3 - (4.625 + 0.5) = 3 - (\frac{1}{2} + 2 + \frac{1}{8} + 2) = 3 - (\frac{4}{8} + \frac{16}{8} + \frac{1}{8} + \frac{16}{8}) = 3 - \frac{37}{8} = \frac{24 - 37}{8} = -\frac{13}{8}$

(4)

まず、$\frac{x^2 + \sqrt{x^3} - \sqrt{x}}{x} = \frac{x^2}{x} + \frac{x^{3/2}}{x} - \frac{x^{1/2}}{x} = x + x^{1/2} - x^{-1/2}$ の不定積分を計算します。

$\int (x + x^{1/2} - x^{-1/2}) dx = \frac{x^2}{2} + \frac{x^{3/2}}{3/2} - \frac{x^{1/2}}{1/2} + C = \frac{x^2}{2} + \frac{2}{3}x^{3/2} - 2x^{1/2} + C = \frac{x^2}{2} + \frac{2}{3}\sqrt{x^3} - 2\sqrt{x} + C$

したがって、定積分は

$\int_2^3 \frac{x^2 + \sqrt{x^3} - \sqrt{x}}{x} dx = [\frac{x^2}{2} + \frac{2}{3}\sqrt{x^3} - 2\sqrt{x}]_2^3 = (\frac{3^2}{2} + \frac{2}{3}\sqrt{3^3} - 2\sqrt{3}) - (\frac{2^2}{2} + \frac{2}{3}\sqrt{2^3} - 2\sqrt{2}) = (\frac{9}{2} + \frac{2}{3} \cdot 3\sqrt{3} - 2\sqrt{3}) - (2 + \frac{2}{3} \cdot 2\sqrt{2} - 2\sqrt{2}) = (\frac{9}{2} + 2\sqrt{3} - 2\sqrt{3}) - (2 + \frac{4}{3}\sqrt{2} - \frac{6}{3}\sqrt{2}) = \frac{9}{2} - (2 - \frac{2}{3}\sqrt{2}) = \frac{9}{2} - 2 + \frac{2}{3}\sqrt{2} = \frac{5}{2} + \frac{2}{3}\sqrt{2}$

3. 最終的な答え

(1) $\frac{187}{2}$

(2) $191 + \sqrt{2}$

(3) $-\frac{13}{8}$

(4) $\frac{5}{2} + \frac{2\sqrt{2}}{3}$