与えられた7つの定積分を計算する問題です。 (1) $\int_{0}^{1} x^2 e^x dx$ (2) $\int_{1}^{2} (x-1)^2 e^{2x} dx$ (3) $\int_{0}^{1} x(x-1)^4 dx$ (4) $\int_{0}^{\frac{\pi}{2}} \sin^4 x dx$ (5) $\int_{1}^{2} (\log x)^3 dx$ (6) $\int_{1}^{2} \frac{\log x}{\sqrt{x}} dx$ (7) $\int_{0}^{1} x \arctan x dx$

1. 問題の内容

与えられた7つの定積分を計算する問題です。

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

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

(3) $\int_{0}^{1} x(x-1)^4 dx$

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

(5) $\int_{1}^{2} (\log x)^3 dx$

(6) $\int_{1}^{2} \frac{\log x}{\sqrt{x}} dx$

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

2. 解き方の手順

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

部分積分を2回行う。

$u = x^2$, $dv = e^x dx$とすると、$du = 2x dx$, $v = e^x$。

$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$

さらに、$u = 2x$, $dv = e^x dx$とすると、$du = 2 dx$, $v = e^x$。

$\int 2x e^x dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2 e^x$

したがって、$\int x^2 e^x dx = x^2 e^x - (2x e^x - 2 e^x) = x^2 e^x - 2x e^x + 2 e^x$

$\int_{0}^{1} x^2 e^x dx = [x^2 e^x - 2x e^x + 2 e^x]_{0}^{1} = (e - 2e + 2e) - (0 - 0 + 2) = e - 2$

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

$t=x-1$とおくと，$x=t+1$，$dx=dt$。積分範囲は$0$から$1$。

$\int_{0}^{1} t^2 e^{2(t+1)} dt = e^2 \int_{0}^{1} t^2 e^{2t} dt$

$u=t^2$, $dv=e^{2t} dt$とすると、$du=2t dt$, $v=\frac{1}{2}e^{2t}$。

$\int t^2 e^{2t} dt = \frac{1}{2}t^2 e^{2t} - \int t e^{2t} dt$

$u=t$, $dv=e^{2t} dt$とすると、$du=dt$, $v=\frac{1}{2}e^{2t}$。

$\int t e^{2t} dt = \frac{1}{2}t e^{2t} - \int \frac{1}{2}e^{2t} dt = \frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t}$

$\int t^2 e^{2t} dt = \frac{1}{2}t^2 e^{2t} - (\frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t}) = \frac{1}{2}t^2 e^{2t} - \frac{1}{2}t e^{2t} + \frac{1}{4}e^{2t}$

$e^2 \int_{0}^{1} t^2 e^{2t} dt = e^2 [\frac{1}{2}t^2 e^{2t} - \frac{1}{2}t e^{2t} + \frac{1}{4}e^{2t}]_{0}^{1} = e^2 (\frac{1}{2}e^2 - \frac{1}{2}e^2 + \frac{1}{4}e^2 - \frac{1}{4}) = \frac{e^4 - e^2}{4}$

(3) $\int_{0}^{1} x(x-1)^4 dx = \int_0^1 x(x^4 - 4x^3 + 6x^2 - 4x + 1) dx = \int_0^1 x^5 - 4x^4 + 6x^3 - 4x^2 + x dx = [\frac{x^6}{6} - \frac{4x^5}{5} + \frac{6x^4}{4} - \frac{4x^3}{3} + \frac{x^2}{2}]_0^1 = \frac{1}{6} - \frac{4}{5} + \frac{3}{2} - \frac{4}{3} + \frac{1}{2} = \frac{5 - 24 + 45 - 40 + 15}{30} = \frac{1}{30}(-24-40+45+20) = \frac{1}{30}$

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

$\sin^2 x = \frac{1 - \cos 2x}{2}$より、$\sin^4 x = (\frac{1 - \cos 2x}{2})^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4} = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 - 4\cos 2x + \cos 4x}{8}$

$\int_{0}^{\frac{\pi}{2}} \sin^4 x dx = \int_{0}^{\frac{\pi}{2}} \frac{3 - 4\cos 2x + \cos 4x}{8} dx = \frac{1}{8} [3x - 2\sin 2x + \frac{1}{4}\sin 4x]_{0}^{\frac{\pi}{2}} = \frac{1}{8}(\frac{3\pi}{2}) = \frac{3\pi}{16}$

(5) $\int_{1}^{2} (\log x)^3 dx$

$u = (\log x)^3$, $dv = dx$, $du = 3 (\log x)^2 \frac{1}{x} dx$, $v = x$

$\int (\log x)^3 dx = x (\log x)^3 - \int 3 (\log x)^2 dx$

$u = 3(\log x)^2$, $dv = dx$, $du = 6 \log x \frac{1}{x} dx$, $v = x$

$\int 3(\log x)^2 dx = 3x (\log x)^2 - \int 6 \log x dx$

$u = 6 \log x$, $dv = dx$, $du = \frac{6}{x} dx$, $v = x$

$\int 6 \log x dx = 6x \log x - \int 6 dx = 6x \log x - 6x$

$\int (\log x)^3 dx = x (\log x)^3 - 3x (\log x)^2 + 6x \log x - 6x$

$\int_{1}^{2} (\log x)^3 dx = [x (\log x)^3 - 3x (\log x)^2 + 6x \log x - 6x]_{1}^{2} = (2 (\log 2)^3 - 6 (\log 2)^2 + 12 \log 2 - 12) - (0 - 0 + 0 - 6) = 2 (\log 2)^3 - 6 (\log 2)^2 + 12 \log 2 - 6$

(6) $\int_{1}^{2} \frac{\log x}{\sqrt{x}} dx$

$u = \log x$, $dv = \frac{1}{\sqrt{x}} dx$, $du = \frac{1}{x} dx$, $v = 2\sqrt{x}$

$\int \frac{\log x}{\sqrt{x}} dx = 2\sqrt{x} \log x - \int \frac{2\sqrt{x}}{x} dx = 2\sqrt{x} \log x - \int \frac{2}{\sqrt{x}} dx = 2\sqrt{x} \log x - 4\sqrt{x}$

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

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

$u = \arctan x$, $dv = x dx$, $du = \frac{1}{1+x^2} dx$, $v = \frac{x^2}{2}$

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

$\int \frac{x^2}{1+x^2} dx = \int \frac{1+x^2 - 1}{1+x^2} dx = \int 1 - \frac{1}{1+x^2} dx = x - \arctan x$

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

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

3. 最終的な答え

(1) $e-2$

(2) $\frac{e^4 - e^2}{4}$

(3) $\frac{1}{30}$

(4) $\frac{3\pi}{16}$

(5) $2 (\log 2)^3 - 6 (\log 2)^2 + 12 \log 2 - 6$

(6) $2\sqrt{2} \log 2 - 4\sqrt{2} + 4$

(7) $\frac{\pi}{4} - \frac{1}{2}$