第10章 10.1

(1)

$\int_0^2 x^2 dx = \left[\frac{1}{3}x^3\right]_0^2 = \frac{8}{3}$

(2)

$y=0$ のとき $x^3-1=(x-1)(x^2+x+1)=0$ $x=1$ これから

$\int_1^2 (x^3-1)dx = \left[\frac{1}{4}x^4-x\right]_1^2 = 4-2-\frac{1}{4}+1=\frac{11}{4}$

(3)

$y=0$ のとき $x^2(3-x)=0$ $x=0,3$

$\int_0^3 (3x^2-x^3)dx = \left[x^3-\frac{1}{4}x^4\right]_0^3 = 27-\frac{81}{4}=\frac{27}{4}$

(4)

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

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

(5)

$\int_0^{\frac{\pi}{4}} \tan x\,dx$ $\cos x = t$ とおくと $\sin x\,dx = dt$ $x:0\to\dfrac{\pi}{4}$ $t:0\to\dfrac{1}{\sqrt2}$ これから

$\int_0^{\frac{1}{\sqrt2}} -\frac{1}{t} = \left[\log|t|\right]_0^{\frac{1}{\sqrt2}} = -\log\frac{1}{\sqrt2} = \frac{1}{2}\log 2$

(6)

$\int_1^e \log x\,dx = \left[x\log x - x\right]_1^e = e\log e - e - 0 + 1 = 1$