次の定積分を求めます。 (1) $\int_{1}^{e} \frac{1}{x} dx$ (2) $\int_{1}^{2} \frac{1}{y^3} dy$ (3) $\int_{1}^{2} (x^3 - 3x^2 + \frac{1}{\sqrt{x}}) dx$ (4) $\int_{0}^{\pi} (2\sin{x} + 3\cos{x}) dx$ (5) $\int_{0}^{1} e^x dx$ (6) $\int_{0}^{\log{a}} e^x dx$

1. 問題の内容

次の定積分を求めます。

(1) $\int_{1}^{e} \frac{1}{x} dx$

(2) $\int_{1}^{2} \frac{1}{y^3} dy$

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

(4) $\int_{0}^{\pi} (2\sin{x} + 3\cos{x}) dx$

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

(6) $\int_{0}^{\log{a}} e^x dx$

2. 解き方の手順

(1)

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

(2)

$\int_{1}^{2} \frac{1}{y^3} dy = \int_{1}^{2} y^{-3} dy = [\frac{y^{-2}}{-2}]_{1}^{2} = [-\frac{1}{2y^2}]_{1}^{2} = -\frac{1}{2(2^2)} - (-\frac{1}{2(1^2)}) = -\frac{1}{8} + \frac{1}{2} = \frac{-1 + 4}{8} = \frac{3}{8}$

(3)

$\int_{1}^{2} (x^3 - 3x^2 + \frac{1}{\sqrt{x}}) dx = \int_{1}^{2} (x^3 - 3x^2 + x^{-\frac{1}{2}}) dx = [\frac{x^4}{4} - x^3 + 2x^{\frac{1}{2}}]_{1}^{2} = (\frac{2^4}{4} - 2^3 + 2\sqrt{2}) - (\frac{1^4}{4} - 1^3 + 2\sqrt{1}) = (4 - 8 + 2\sqrt{2}) - (\frac{1}{4} - 1 + 2) = -4 + 2\sqrt{2} - \frac{1}{4} + 1 - 2 = -5 - \frac{1}{4} + 2\sqrt{2} = -\frac{21}{4} + 2\sqrt{2}$

(4)

$\int_{0}^{\pi} (2\sin{x} + 3\cos{x}) dx = [-2\cos{x} + 3\sin{x}]_{0}^{\pi} = (-2\cos{\pi} + 3\sin{\pi}) - (-2\cos{0} + 3\sin{0}) = (-2(-1) + 3(0)) - (-2(1) + 3(0)) = 2 - (-2) = 4$

(5)

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

(6)

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

3. 最終的な答え

(1) 1

(2) 3/8

(3) $-\frac{21}{4} + 2\sqrt{2}$

(4) 4

(5) $e-1$

(6) $a-1$