3.7 定積分

1 区間

を

等分する．

軸上の点 $\frac{i}{n}$ に対応する値 $(\frac{i}{n})^2$ を高さとし，底辺が $\frac{i}{n} - \frac{i-1}{n} = \frac{1}{n}$ を底辺とする長方形を考える．この長方形を $x = 0$

から

までの間で加えると，Riemann和とよばれる次の和を得る．

$\displaystyle \sum_{i=1}^{n} \frac{1}{n}\left(\frac{i}{n}\right) = \frac{1}{n}\sum_{i=1}^{n}\left(\frac{i}{n}\right)$

$\displaystyle \int_{0}^{1}x^2 \; dx$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\left(\frac{i}{n}\right)$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{1}{n}\cdot \frac{1}{n^2}\sum_{i=1}^{n}i^2$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty} \frac{1}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty} \frac{2n^3 + 3n^2 + n}{6n^3} = \frac{1}{3}$

とすると $\displaystyle{\int_{a}^{b}f(x)\; dx = [F(x)]_{a}^{b} = F(a) - F(b)}$

$\displaystyle \int_{0}^{1}(x^2 + 3)\; dx = \left[\frac{x^3}{3} + 3x\right]_{0}^{1} = \frac{1}{3} + 3 = \frac{10}{3}$

$\displaystyle \int_{1}^{2}\frac{x^2 - 1}{x}\; dx$	$\displaystyle =$	$\displaystyle \int_{1}^{2}(x - \frac{1}{x})\; dx$
	$\displaystyle =$	$\displaystyle \left[\frac{x^2}{2} - \log\vert x\vert\right]_{1}^{2}$
	$\displaystyle =$	$\displaystyle 2 - \log{2} - \frac{1}{2} = \frac{3}{2} - \log{2}$

$\displaystyle \int_{0}^{1}\sqrt{x^3}\; dx = \int_{0}^{1}x^{\frac{3}{2}}\; dx = \left[\frac{2}{5}x^{\frac{5}{2}}\right]_{0}^{1} = \frac{2}{5}$

$\displaystyle \int_{0}^{\pi}\cos{x}\; dx = [\sin{x}]_{0}^{\pi} = 0$

$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin{x}\; dx = [-\cos{x}]_{0}^{\frac{\pi}{2}} = \cos{0} = 1$

$\displaystyle \int_{a}^{c}f(x)\; dx + \int_{c}^{b}f(x)\;dx = \int_{a}^{b}f(x)\;dx$

$\displaystyle \int_{0}^{5}f(x)\; dx = \int_{0}^{2}f(x)\; dx + \int_{2}^{5}f(x)\; dx = 4 + 1 = 5$

$\displaystyle \int_{1}^{2}f(x)\; dx = \int_{0}^{2}f(x)\; dx - \int_{0}^{1}f(x)\; dx = 4 - 6 = -2$

$\displaystyle \int_{1}^{5}f(x)\; dx = \int_{1}^{2}f(x)\; dx + \int_{2}^{5}f(x)\; dx = -2 + 1 = -1$

$\displaystyle \int_{0}^{0}f(x)\; dx = 0$

$\displaystyle \int_{2}^{0}f(x)\; dx = -\int_{0}^{2}f(x)\; dx = -4$

$\displaystyle \frac{d}{dx}\int_{a}^{x}f(t)\;dt = f(x)$

$\displaystyle \frac{d}{dx}\int_{1}^{x}\sin{t}\;dt = \sin{x}$

$\displaystyle \frac{d}{dx}\int_{x}^{1}\cos{t}\;dt = -\frac{d}{dx}\int_{1}^{x}\cos{t}\;dt = -\cos{x}$

$\displaystyle \frac{d}{dx}\int_{0}^{2x}\sqrt{\sin{t}}\;dt$	$\displaystyle =$	$\displaystyle \frac{d}{dx}\int_{a}^{u}\sqrt{\sin{t}}\;dt$
	$\displaystyle =$	$\displaystyle \frac{d}{du}(\int_{a}^{u}\sqrt{\sin{t}}\;dt) \frac{du}{dx}$
	$\displaystyle =$	$\displaystyle \sqrt{\sin{u}} \cdot 2 = 2\sqrt{\sin{2x}}$

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}f(\frac{i}{n}) = \int_{0}^{1}f(x)\; dx$

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\frac{i}{n} = \int_{0}^{1}x\;dx = [\frac{x^2}{2}]_{0}^{1} = \frac{1}{2}$

		$\displaystyle \lim_{n \to \infty}\frac{1}{n}\left(\frac{1}{2 + \frac{1}{n}} + \frac{1}{2 + \frac{2}{n}} + \cdots + \frac{1}{2 + \frac{n}{n}}\right)$
	$\displaystyle =$	$\displaystyle \int_{0}^{1}\frac{1}{2+x}\;dx = [\log\vert 2+x\vert]_{0}^{1}$
	$\displaystyle =$	$\displaystyle \log{3} - \log{2} = \log{\frac{3}{2}}$

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sqrt{\frac{i}{n}} = \int_{0}^{1}\sqrt{x}\;dx = [\frac{2}{3}x^{3/2}]_{0}^{1} = \frac{2}{3}$