第9章 9.11

(1)

$\int_0^3 \sqrt{3-t}\,dt = -\frac{2}{3}\Big[(3-t)^{\frac{3}{2}}\Big]_0^3 = -\frac{2}{3}\Big\{(3-3)^{\frac{3}{2}}-(3-0)^{\frac{3}{2}}\Big\} = -\frac{2}{3}\cdot 3\sqrt{3} = 2\sqrt{3}$

(2)

$\int_{-1}^1 x^2\sqrt{1+x^3}\,dx$

$1+x^3=t$ とおくと $3x^2dx=dt$ x: $-1\sim1$ t: $0\sim2$ これから

$=\int_0^2 \frac{1}{3}\sqrt{t}\,dt = \frac{1}{3}\cdot\frac{2}{3}\Big[t^{\frac{3}{2}}\Big]_0^2 = \frac{2}{9}\left(2^{\frac{3}{2}}-0\right) = \frac{4\sqrt{2}}{9}$

(3)

$\int_1^2 x\cdot \log x\,dx$

$\int x\log x\,dx = I$ とおくと

$I = x(x\log x - x) - \int (x\log x - x)\,dx = x^2\log x - x^2 - \int x\log x\,dx + \int x\,dx = x^2\log x - x^2 - I + \frac{1}{2}x^2$

$2I = x^2\left(\log x - \frac{1}{2}\right) \qquad I = \frac{1}{2}x^2\left(\log x - \frac{1}{2}\right) \text{となるので}$

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

(4)

$\int_e^{e^2} \frac{dx}{x\cdot \log x}$

$\log x = t$ とおくと $\dfrac{1}{x}dx=dt$ x: $e \sim e^2$ t: $1\sim2$ これから

$=\int_1^2 \frac{1}{t}\,dt = \big[\log t\big]_1^2 = \log 2 - \log 1 = \log 2$