第10章 10.2

(1)

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

(2)

$y=0$ のとき $x^2(x-1)=0$ $x=0,1$

$\int_{0}^{1}(x^3-x^2)dx=\left[\frac{1}{4}x^4-\frac{1}{3}x^3\right]_{0}^{1}=\frac{1}{4}-\frac{1}{3}=-\frac{1}{12}$ 面積なので $\dfrac{1}{12}$

(3)

$1-x=2x^2$ $2x^2+x-1=(2x-1)(x+1)=0$ $x=-1,\dfrac{1}{2}$

$\int_{-1}^{\frac{1}{2}}(1-x)dx-\int_{-1}^{\frac{1}{2}}2x^2dx=\left[x-\frac{1}{2}x^2\right]_{-1}^{\frac{1}{2}}-\left[\frac{2}{3}x^3\right]_{-1}^{\frac{1}{2}}=\frac{1}{2}-\frac{1}{8}+1+\frac{1}{2}-\frac{1}{12}-\frac{2}{3}=\frac{27}{24}=\frac{9}{8}$

(4)

$y=0$ のとき $x(x^2-1)=x(x-1)(x+1)=0$ $x=0,-1,1$

$\int_{-1}^{0}x(x^2-1)dx-\int_{0}^{1}x(x^2-1)dx=\int_{-1}^{0}(x^3-x)dx-\int_{0}^{1}(x^3-x)dx=\left[\frac{1}{4}x^4-\frac{1}{2}x^2\right]_{-1}^{0}-\left[\frac{1}{4}x^4-\frac{1}{2}x^2\right]_{0}^{1}$

$=-\frac{1}{4}+\frac{1}{2}-\frac{1}{4}+\frac{1}{2}=\frac{1}{2}$

(5)

$y=0$ のとき $x=0$

$\int_{0}^{2}\frac{x}{x^2+4}dx$ $x^2=t$ とおくと $2xdx=dt$ $x:0\to2$ $t:0\to4$ これから

$\int_{0}^{4}\frac{1}{2}\cdot\frac{1}{t+4}dt=\frac{1}{2}\left[\log|t+4|\right]_{0}^{4}=\frac{1}{2}(\log8-\log4)=\frac{1}{2}\log2$

(6)

$y = (1 - \sqrt{x})^2$

定義域は $x, y > 0$ $y = 0$ のとき $x = 1$ これから

$\int_0^1 (1 - \sqrt{x})^2\, dx = \int_0^1 (1 - 2\sqrt{x} + x)\, dx = \left[ x - \frac{4}{3}\sqrt{x^3} + \frac{1}{2}x^2 \right]_0^1 = 1 - \frac{4}{3} + \frac{1}{2} = \frac{1}{6}$