第10章 10.10

(1)

$x(3-x)=2$ $\quad$ $x^2-3x+2=(x-2)(x-1)=0$ $\quad$ $x=1,2$

これから

$\int_1^2 (3-x)-\frac{2}{x}\,dx=\left[3x-\frac{1}{2}x^2-2\log x\right]_1^2=6-2-2\log 2-3+\frac{1}{2}-0=\frac{3}{2}-2\log 2$

(2)

$2\left\{\int_1^2(\sqrt{x}-(2-x))dx+\int_2^4(\sqrt{x}-(x-2))dx\right\}=2\left(\left[\frac{2}{3}\sqrt{x^3}-2x+\frac{1}{2}x^2\right]_1^2+\left[\frac{2}{3}\sqrt{x^3}-\frac{1}{2}x^2+2x\right]_2^4\right)$

$=2\left(\frac{4\sqrt{2}}{3}-4+2-\frac{2}{3}+2-\frac{1}{2}+\frac{16}{3}-8+8-\frac{4\sqrt{2}}{3}+2-4\right)=2\cdot\frac{13}{6}=\frac{13}{3}$

(3)

$x^2+(x^2+2-2)^2=2$ $\quad$ $x^4+x^2-2=0$ $\quad$ $(x^2+2)(x^2-1)=(x^2+2)(x-1)(x+1)=0$ $\quad$ $x=1$

これから

$2\int_0^1\{(\sqrt{2-x^2}+2)-(x^2+2)\}dx=2\int_0^1\sqrt{2-x^2}\,dx+2\int_0^1(2-x^2-2)dx$

第１項で $x=\sqrt{2}\sin t$ とおくと $\quad$ $dx=\sqrt{2}\cos t\,dt$ $\quad$ $x:0\to 1$ $\quad$ $t:0\to\dfrac{\pi}{4}$ $\quad$ これから

$=2\int_0^{\frac{\pi}{4}}\sqrt{2-2\sin^2 t}\sqrt{2}\cos t\,dt-2\int_0^1 x^2dx=4\int_0^{\frac{\pi}{4}}\cos^2 t\,dt-2\left[\frac{1}{3}x^3\right]_0^1=2\int_0^{\frac{\pi}{4}}(1+\cos 2t)dt-\frac{2}{3}$

$=2\left[t+\frac{1}{2}\sin 2t\right]_0^{\frac{\pi}{4}}-\frac{2}{3}=2\left(\frac{\pi}{4}+\frac{1}{2}\right)-\frac{2}{3}=\frac{\pi}{2}+\frac{1}{3}$