与えられた9つの定積分をそれぞれ計算します。 (1) $\int_{1}^{4} \frac{1}{\sqrt{x}} dx$ (2) $\int_{0}^{\pi} \cos{t} dt$ (3) $\int_{1}^{e} \frac{1}{u} du$ (4) $\int_{0}^{1} \sqrt[3]{x^2} dx$ (5) $\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2{x}} dx$ (6) $\int_{-1}^{0} \frac{1}{e^x} dx$ (7) $\int_{0}^{1} e^{x-1} dx$ (8) $\int_{1}^{2} 2^x dx$ (9) $\int_{0}^{\frac{\pi}{2}} \sin^2{x} dx$

1. 問題の内容

与えられた9つの定積分をそれぞれ計算します。

(1) $\int_{1}^{4} \frac{1}{\sqrt{x}} dx$

(2) $\int_{0}^{\pi} \cos{t} dt$

(3) $\int_{1}^{e} \frac{1}{u} du$

(4) $\int_{0}^{1} \sqrt[3]{x^2} dx$

(5) $\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2{x}} dx$

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

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

(8) $\int_{1}^{2} 2^x dx$

(9) $\int_{0}^{\frac{\pi}{2}} \sin^2{x} dx$

2. 解き方の手順

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

(2) $\int_{0}^{\pi} \cos{t} dt = [\sin{t}]_{0}^{\pi} = \sin{\pi} - \sin{0} = 0 - 0 = 0$

(3) $\int_{1}^{e} \frac{1}{u} du = [\ln{|u|}]_{1}^{e} = \ln{e} - \ln{1} = 1 - 0 = 1$

(4) $\int_{0}^{1} \sqrt[3]{x^2} dx = \int_{0}^{1} x^{\frac{2}{3}} dx = [\frac{3}{5}x^{\frac{5}{3}}]_{0}^{1} = \frac{3}{5}(1)^{\frac{5}{3}} - \frac{3}{5}(0)^{\frac{5}{3}} = \frac{3}{5} - 0 = \frac{3}{5}$

(5) $\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2{x}} dx = \int_{0}^{\frac{\pi}{4}} \sec^2{x} dx = [\tan{x}]_{0}^{\frac{\pi}{4}} = \tan{\frac{\pi}{4}} - \tan{0} = 1 - 0 = 1$

(6) $\int_{-1}^{0} \frac{1}{e^x} dx = \int_{-1}^{0} e^{-x} dx = [-e^{-x}]_{-1}^{0} = -e^{-0} - (-e^{-(-1)}) = -e^0 + e^1 = -1 + e = e-1$

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

(8) $\int_{1}^{2} 2^x dx = [\frac{2^x}{\ln{2}}]_{1}^{2} = \frac{2^2}{\ln{2}} - \frac{2^1}{\ln{2}} = \frac{4}{\ln{2}} - \frac{2}{\ln{2}} = \frac{2}{\ln{2}}$

(9) $\int_{0}^{\frac{\pi}{2}} \sin^2{x} dx$

$\sin^2{x} = \frac{1-\cos{2x}}{2}$

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

3. 最終的な答え

(1) 2

(2) 0

(3) 1

(4) 3/5

(5) 1

(6) e-1

(7) 1 - 1/e

(8) 2/ln(2)

(9) π/4