第10章 10.9

(1)

$\left(\dfrac{x^2}{4a}\right)^2=4ax$ $\qquad$ $x\left(\dfrac{1}{16a^2}x^3-4a\right)=0$ $\qquad$ $x=0,4a$ これから

$S=\int_0^{4a}\left(\sqrt{4ax}-\frac{x^2}{4a}\right)dx=\left[\frac{2}{3}\sqrt{4ax^3}-\frac{1}{12a}x^3\right]_0^{4a}=\left(\frac{2}{3}\cdot16a^2-\frac{16}{3}a^2\right)=\frac{16}{3}a^2$

(2)

$4(x+1)=8(2-x)$ $\qquad$ $12x=12$ $\qquad$ $x=1$ これから

$S=\int_{-1}^{1}\sqrt{4(x+1)}\,dx+\int_{1}^{2}\sqrt{8(2-x)}\,dx=\left[\frac{2}{12}\sqrt{\{4(x+1)\}^3}\right]_{-1}^{1}+\left[-\frac{2}{24}\sqrt{\{8(2-x)\}^3}\right]_{1}^{2}$

$=\frac{16\sqrt2}{6}+\frac{16\sqrt2}{12}=\frac{32\sqrt2+16\sqrt2}{12}=4\sqrt2$

(3)

$x^2+x^4=2x$ $\qquad$ $x(x^3+x-2)=x(x-1)(x^2+x+2)=0$ $\qquad$ $x=0,1$ これから

$S=\int_0^1\sqrt{2x-x^2}-x^2\,dx=\int_0^1\sqrt{1-(x-1)^2}-x^2\,dx=\int_0^1\sqrt{1-(x-1)^2}\,dx-\left[\frac{1}{3}x^3\right]_0^1=\int_0^1\sqrt{1-(x-1)^2}\,dx-\frac{1}{3}$

$x-1=\sin t$ とおくと $dx=\cos t\,dt$ $x:0\to1$ $t:-\dfrac{\pi}{2}\to0$ これから

$S=\int_{-\frac12\pi}^{0}\sqrt{1-\sin^2t}\cdot\cos t\,dt-\frac13=\int_{-\frac12\pi}^0\cos^2t\,dt-\frac13=\int_{-\frac12\pi}^0\frac{1+\cos2x}{2}\,dt-\frac13=\left[\frac12x+\frac14\sin2x\right]_{-\frac12\pi}^0-\frac13$

$=\frac14\pi-\frac13$

(4)

$\sqrt{x}+x=2$ $\qquad$ $x+\sqrt{x}-2=0$ $\qquad$ $(\sqrt{x}-1)(\sqrt{x}+2)=0$ $\qquad$ $x=1$ これから

$S=\int_0^1(2-\sqrt{x})^2-x^2\,dx=\int_0^1(4-4\sqrt{x}+x-x^2)\,dx=\left[4x-\frac83\sqrt{x^3}+\frac12x^2-\frac13x^3\right]_0^1=4-\frac83+\frac12-\frac13=\frac32$