第5章 5.8

(1)

$y = \frac{x^3}{3} + \frac{1}{4x} \qquad [1,2]$

$1+\left(\frac{dy}{dx}\right)^2 = 1+\left(x^2-\frac{1}{4x^2}\right)^2 = 1+x^4-\frac{1}{2}+\frac{1}{16x^4} = \frac{16x^8+1+8x^4}{16x^4} = \frac{\left(x^4+\dfrac{1}{4}\right)^2}{x^4}$

$S = \int_1^2 \sqrt{\frac{\left(x^4+\dfrac{1}{4}\right)^2}{x^4}}\,dx = \int_1^2 \frac{x^4+\dfrac{1}{4}}{x^2}\,dx = \int_1^2 x^2+\frac{1}{4x^2}\,dx = \left[\frac{1}{3}x^3-\frac{1}{4x}\right]_1^2 = \frac{8}{3}-\frac{1}{8}-\frac{1}{3}+\frac{1}{4} = \frac{59}{24}$

(2)

$y = x^2-1$

$1+\left(\frac{dy}{dx}\right)^2 = 1+(2x)^2 = 1+4x^2$

$S = \int_{-1}^1 \sqrt{1+4x^2}\,dx \qquad 2x=t \quad \text{とおくと} \quad 2dx=dt \qquad x:-1\to 1 \qquad t:-2\to 2 \quad \text{これから}$

$= \int_{-2}^2 \sqrt{1+t^2}\cdot\frac{1}{2}\,dt = \left[\frac{1}{2}\left\{t\sqrt{1+t^2}+\log\left|t+\sqrt{1+t^2}\right|\right\}\right]_0^2 = \frac{1}{2}\left\{2\sqrt5+\log(2+\sqrt5)\right\}$

$= \sqrt5+\frac{1}{2}\log(2+\sqrt5)$

(3)

$\sqrt x + \sqrt y = 1$

$y = (1-\sqrt x)^2 \qquad y' = 2(1-\sqrt x)\cdot\left(-\frac{1}{2\sqrt x}\right) = \frac{\sqrt x-1}{\sqrt x}$

$1+\left(\frac{dy}{dx}\right)^2 = 1+\left(\frac{\sqrt x-1}{\sqrt x}\right)^2 = \frac{x+x-2\sqrt x+1}{x} = \frac{2x-2\sqrt x+1}{x}$

$S = \int_0^1 \sqrt{\frac{2x-2\sqrt x+1}{x}}\,dx \qquad \sqrt x=t \quad \text{とおくと} \quad \frac{1}{2\sqrt x}dx=dt \qquad x:0\to 1 \qquad t:0\to 1 \quad \text{これから}$

$= \int_0^1 \sqrt{2t^2-2t+1}\cdot 2\,dt = \int_0^1 2\sqrt2\cdot\sqrt{\left(t-\frac{1}{2}\right)^2+\frac{1}{4}}\,dt$

$= 2\sqrt{2}\left[\frac{1}{2}\left\{t\sqrt{\left(t-\frac{1}{2}\right)^2+\frac{1}{4}}+\frac{1}{4}\log\left|t-\frac{1}{2}+\sqrt{\left(t-\frac{1}{2}\right)^2+\frac{1}{4}}\right|\right\}\right]_0^1 = 1+\frac{\sqrt{2}}{4}\log\frac{\dfrac{1}{2}+\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{2}+\dfrac{1}{\sqrt{2}}}$

$= 1+\frac{1}{\sqrt{2}}\log(1+\sqrt{2})$