早稲田大学 創造理工学研究科 経営システム工学専攻 2021年8月実施 微积分

标签：

• Waseda-University
• Definite-Integral
• Integration-by-Substitution
• Integration

Author

思齐塾, 祭音Myyura

Description

次の定積分について、

$$\int_0^2 \frac{1}{\sqrt{x^2 + 1}} dx$$

$x = \frac{1}{2}(y - \frac{1}{y}), (y > 0)$ と置換(substitute) して計算せよ。

题目描述

对定积分

$$\int_0^2\frac{1}{\sqrt{x^2+1}}\,dx,$$

使用指定代换

$$x=\frac12\left(y-\frac1y\right),\qquad y>0,$$

完成计算。

Kai

Let $x = \frac{1}{2}(y - \frac{1}{y})$ . Then $dx = \frac{1}{2}(1 + \frac{1}{y^2}) dy = \frac{1}{2}(\frac{y^2 + 1}{y^2}) dy$ . Also, $x^2 = \frac{1}{4}(y^2 - 2 + \frac{1}{y^2})$ . Then, $x^2 + 1 = \frac{1}{4}(y^2 - 2 + \frac{1}{y^2}) + 1 = \frac{1}{4}(y^2 + 2 + \frac{1}{y^2}) = \frac{1}{4}(y + \frac{1}{y})^2$ . So, $\sqrt{x^2 + 1} = \frac{1}{2}(y + \frac{1}{y})$ . Then, $\frac{1}{\sqrt{x^2 + 1}} = \frac{2}{y + \frac{1}{y}} = \frac{2y}{y^2 + 1}$ . So, $\frac{1}{\sqrt{x^2 + 1}} dx = \frac{2y}{y^2 + 1} \cdot \frac{1}{2} \frac{y^2 + 1}{y^2} dy = \frac{1}{y} dy$ .

Now, when $x=0$ , $\frac{1}{2}(y - \frac{1}{y}) = 0$ , which implies $y - \frac{1}{y} = 0$ , so $y^2 = 1$ . Since $y > 0$ , $y=1$ . When $x=2$ , $\frac{1}{2}(y - \frac{1}{y}) = 2$ , which implies $y - \frac{1}{y} = 4$ , so $y^2 - 4y - 1 = 0$ . Then $y = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$ . Since $y > 0$ , $y = 2 + \sqrt{5}$ . Therefore, the integral becomes:

$$\int_1^{2+\sqrt{5}} \frac{1}{y} dy = \left[ \ln y \right]_1^{2+\sqrt{5}} = \ln(2+\sqrt{5}) - \ln(1) = \ln(2+\sqrt{5})$$

So, the answer is $\ln(2+\sqrt{5})$ .