第9章 9.2

(1)

$\int_1^2 (2x-1)^{\frac{3}{2}}dx = \left[\frac{2}{5}(2x-1)^{\frac{5}{2}}\cdot\frac{1}{2}\right]_1^2 = \frac{1}{5}\left((4-1)^{\frac{5}{2}}-(2-1)^{\frac{5}{2}}\right) = \frac{1}{5}(9\sqrt{3}-1)$

(2)

$\int_1^3 (x-1)^5 dx = \left[\frac{1}{6}(x-1)^6\right]_1^3 = \frac{1}{6}((3-1)^6-(1-1)^6) = \frac{32}{3}$

(3)

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

(4)

$\int_0^1 \frac{dt}{t+1} = [\log|t+1|]_0^1 = \log 2 - \log 1 = \log 2$

(5)

$\int_1^2 e^{3t}dt = \left[\frac{1}{3}e^{3t}\right]_1^2 = \frac{1}{3}(e^6-e^3) = \frac{1}{3}e^3(e^3-1)$

(6)

$\int_1^2 x\sqrt{x-1}\,dx = \left[x\cdot\frac{2}{3}\sqrt{(x-1)^3}\right]_1^2 - \frac{2}{3}\int_1^2 \sqrt{(x-1)^3}dx = \frac{2}{3}\left(2\cdot\sqrt{(2-1)^3} - 1\cdot\sqrt{(1-1)^3}\right) - \frac{2}{3}\int_1^2 \sqrt{(x-1)^3}dx$ $= \frac{4}{3} - \frac{2}{3}\cdot\frac{2}{5}\left[\sqrt{(x-1)^5}\right]_1^2 = \frac{4}{3} - \frac{4}{15} = \frac{16}{15}$

(7)

$\int_0^3 \frac{x}{\sqrt{4-x}}dx = \left[x\cdot(-2)\sqrt{4-x}\right]_0^3 + 2\int_0^3 \sqrt{4-x}\,dx = -6 + 2\cdot\left(-\frac{2}{3}\right)\left[\sqrt{(4-x)^3}\right]_0^3 = -6 - \frac{4}{3}(1-8) = \frac{10}{3}$

(8)

$\int_1^3 \frac{x}{1+x^2}dx$

$1+x^2 = t$ とおくと $2x\,dx = dt$ $x:1\to3$ $t:2\to10$ これより

$= \int_2^{10} \frac{1}{t}\cdot\frac{1}{2}dt = \frac{1}{2}[\log|t|]_2^{10} = \frac{1}{2}(\log 10 - \log 2) = \frac{1}{2}\log 5$

(9)

$\int_0^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}}\,dx$

$1-x^2=t$ とおくと $-2x\,dx=dt$

$x:0\to\dfrac{1}{2}$ $t:1\to\dfrac{3}{4}$ これより

$=\int_1^{\frac{3}{4}} \frac{1}{\sqrt{t}}\cdot\left(-\frac{1}{2}\right)dt=-\frac{1}{2}\left[2\sqrt{t}\right]_1^{\frac{3}{4}}=-\left(\frac{\sqrt{3}}{2}-1\right)=1-\frac{\sqrt{3}}{2}$

(10)

$\int_1^e \frac{\log x}{x}\,dx=\left[\log x\cdot\log x\right]_1^e-\int_1^e \frac{\log x}{x}\,dx$

$2\int_1^e \frac{\log x}{x}\,dx=\left[\log^2 x\right]_1^e=1$

$\int_1^e \frac{\log x}{x}\,dx=\frac{1}{2}$