与えられた5つの定積分の問題を解きます。 [1] $\int_{0}^{1} \frac{e^x}{e^x + 1} dx$ [2] $\int_{0}^{1} |(x-4)(x-1)^3| dx$ [3] $\int_{0}^{\sqrt{3}} \frac{t^2}{(1+t^2)^2} dt$ [4] $\int_{1}^{e} x^2 \log x dx$ [5] $\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$

1. 問題の内容

与えられた5つの定積分の問題を解きます。

[1] $\int_{0}^{1} \frac{e^x}{e^x + 1} dx$

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

[3] $\int_{0}^{\sqrt{3}} \frac{t^2}{(1+t^2)^2} dt$

[4] $\int_{1}^{e} x^2 \log x dx$

[5] $\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$

2. 解き方の手順

[1]

$u = e^x + 1$ と置換すると、$du = e^x dx$ となります。

$x = 0$ のとき、$u = e^0 + 1 = 1 + 1 = 2$

$x = 1$ のとき、$u = e^1 + 1 = e + 1$

したがって、

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

[2]

$f(x) = (x-4)(x-1)^3$ を考えます。$0 \le x \le 1$ の範囲では、$x-4 < 0$ であり、$(x-1)^3 \le 0$ であるため、$(x-4)(x-1)^3 \ge 0$ となります。したがって、絶対値を外すことができます。

$\int_{0}^{1} |(x-4)(x-1)^3| dx = \int_{0}^{1} (x-4)(x-1)^3 dx = \int_{0}^{1} (x-4)(x^3 - 3x^2 + 3x - 1) dx$

$= \int_{0}^{1} (x^4 - 3x^3 + 3x^2 - x - 4x^3 + 12x^2 - 12x + 4) dx = \int_{0}^{1} (x^4 - 7x^3 + 15x^2 - 13x + 4) dx$

$= [\frac{1}{5}x^5 - \frac{7}{4}x^4 + 5x^3 - \frac{13}{2}x^2 + 4x]_{0}^{1} = \frac{1}{5} - \frac{7}{4} + 5 - \frac{13}{2} + 4 = \frac{4 - 35 + 100 - 130 + 80}{20} = \frac{19}{20}$

[3]

$t = \tan \theta$ と置換すると、$dt = \sec^2 \theta d\theta$ となります。

$t = 0$ のとき、$\theta = 0$

$t = \sqrt{3}$ のとき、$\theta = \frac{\pi}{3}$

$\int_{0}^{\sqrt{3}} \frac{t^2}{(1+t^2)^2} dt = \int_{0}^{\frac{\pi}{3}} \frac{\tan^2 \theta}{(1+\tan^2 \theta)^2} \sec^2 \theta d\theta = \int_{0}^{\frac{\pi}{3}} \frac{\tan^2 \theta}{(\sec^2 \theta)^2} \sec^2 \theta d\theta = \int_{0}^{\frac{\pi}{3}} \frac{\tan^2 \theta}{\sec^2 \theta} d\theta$

$= \int_{0}^{\frac{\pi}{3}} \sin^2 \theta d\theta = \int_{0}^{\frac{\pi}{3}} \frac{1 - \cos 2\theta}{2} d\theta = [\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta]_{0}^{\frac{\pi}{3}} = \frac{1}{2} \cdot \frac{\pi}{3} - \frac{1}{4}\sin \frac{2\pi}{3} = \frac{\pi}{6} - \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{6} - \frac{\sqrt{3}}{8}$

[4]

部分積分を行います。$\int x^2 \log x dx$

$u = \log x$, $dv = x^2 dx$ とすると、$du = \frac{1}{x} dx$, $v = \frac{1}{3}x^3$

$\int x^2 \log x dx = \frac{1}{3}x^3 \log x - \int \frac{1}{3}x^3 \cdot \frac{1}{x} dx = \frac{1}{3}x^3 \log x - \frac{1}{3} \int x^2 dx = \frac{1}{3}x^3 \log x - \frac{1}{9}x^3 + C$

$\int_{1}^{e} x^2 \log x dx = [\frac{1}{3}x^3 \log x - \frac{1}{9}x^3]_{1}^{e} = (\frac{1}{3}e^3 \log e - \frac{1}{9}e^3) - (\frac{1}{3} \cdot 1^3 \log 1 - \frac{1}{9} \cdot 1^3) = \frac{1}{3}e^3 - \frac{1}{9}e^3 - 0 + \frac{1}{9} = \frac{2}{9}e^3 + \frac{1}{9}$

[5]

$x = \sin \theta$ と置換すると、$dx = \cos \theta d\theta$ となります。

$x = \frac{1}{\sqrt{2}}$ のとき、$\theta = \frac{\pi}{4}$

$x = \frac{\sqrt{3}}{2}$ のとき、$\theta = \frac{\pi}{3}$

$\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin^2 \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta d\theta = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin^2 \theta}{\cos \theta} \cos \theta d\theta = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \sin^2 \theta d\theta$

$= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1 - \cos 2\theta}{2} d\theta = [\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = (\frac{1}{2} \cdot \frac{\pi}{3} - \frac{1}{4}\sin \frac{2\pi}{3}) - (\frac{1}{2} \cdot \frac{\pi}{4} - \frac{1}{4}\sin \frac{\pi}{2})$

$= \frac{\pi}{6} - \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{\pi}{8} + \frac{1}{4} \cdot 1 = \frac{\pi}{24} - \frac{\sqrt{3}}{8} + \frac{1}{4}$

3. 最終的な答え

[1] $\log(\frac{e+1}{2})$

[2] $\frac{19}{20}$

[3] $\frac{\pi}{6} - \frac{\sqrt{3}}{8}$

[4] $\frac{2e^3+1}{9}$

[5] $\frac{\pi}{24} - \frac{\sqrt{3}}{8} + \frac{1}{4}$