3.7 解答

微分積分学の基本定理は次のように使う． $f(t)$

が

で連続で，

ならば

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

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

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

$\displaystyle \frac{d}{dx}\int_{x}^{x+1}f(t) dt$	$\displaystyle =$	$\displaystyle \frac{d}{dx}[\int_{x}^{a}f(t) dt + \int_{a}^{x+1}f(t) dt]$
	$\displaystyle =$	$\displaystyle -f(x) + f(x+1)\frac{d(x+1)}{dx} = -f(x) + f(x+1)$

$\displaystyle \frac{d}{dx}\int_{0}^{2x}x^{2}f(t) dt$	$\displaystyle =$	$\displaystyle \frac{d}{dx}(x^{2}\int_{0}^{2x}f(t) dt \left(\begin{array}{l} tについての積分なので\\ x^{2}は積分記号の外に\\ 出せる \end{array}\right)$
	$\displaystyle =$	$\displaystyle 2x \int_{0}^{2x}f(t) dt + x^{2}\frac{d}{dx}\int_{0}^{2x}f(t) dt$
	$\displaystyle =$	$\displaystyle 2x \int_{0}^{2x}f(t)dt + x^{2}f(2x)\cdot 2$

2. 定積分は次のようにして求める． $F'(x) = f(x)$

のとき，つまり

が

の原始関数であるとき

$\displaystyle \int_{a}^{b}f(x) dx = F(x)\mid_{a}^{b} = F(b) - F(a)$

$\displaystyle \int_{a}^{b}f(g(x))g'(x) dx$

$\displaystyle \int_{a}^{b}f(g(x))g'(x) dx = \int_{t=g(a)}^{t=g(b)}f(t) dt$

$\begin{displaymath}\begin{array}{llll} x&1 &\Rightarrow & 5\ \hline t&0 &\Rightarrow & 2 \end{array}\end{displaymath}$

$\displaystyle \int_{1}^{5}{2\sqrt{x-1}} dx$	$\displaystyle =$	$\displaystyle 2\int_{0}^{2}{t(2tdt)}$
	$\displaystyle =$	$\displaystyle 4\int_{0}^{2}t^{2} dt = \frac{4}{3}t^{3}\mid_{0}^{2} = \frac{32}{3}$

$\displaystyle \int_{1}^{2}{\frac{2-t}{t^3}} dt$	$\displaystyle =$	$\displaystyle \int_{1}^{2}{(2t^{-3} - t^{-2})} dt = [-t^{-2} + t^{-1}\mid_{1}^{2}$
	$\displaystyle =$	$\displaystyle [-\frac{1}{t^2} + \frac{1}{t}\mid_{1}^{2} = -\frac{1}{4} + \frac{1}{2} - (-1+1) = \frac{1}{4}$

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\cos{x}} dx$

$\displaystyle =$

$\displaystyle \sin{x}\mid_{0}^{\frac{\pi}{2}} = \sin{(\frac{\pi}{2})} - \sin{0} = 1$

$\begin{displaymath}\begin{array}{llll} x&0 & \Rightarrow & 1\ \hline t&0 &\Rightarrow & -1 \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{1}{xe^{-x^2}} dx$	$\displaystyle =$	$\displaystyle \int_{0}^{-1}{e^{t}(-2tdt)}$
	$\displaystyle =$	$\displaystyle 2\int_{-1}^{0}{e^{t}} dt = 2e^{t}\mid_{-1}^{0} = 2(e^{0} - e^{-1}) = 2(1 - \frac{1}{e})$

$\begin{displaymath}\begin{array}{llll} x&0 & \Rightarrow & \log{2}\ \hline t&2 &\Rightarrow & 3 \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{\log{2}}{\frac{e^{x}}{e^{x}+1}} dx$	$\displaystyle =$	$\displaystyle \int_{2}^{3}{\frac{dt}{t}} dt = \frac{1}{a^{2}}\int{\frac{1}{\sec{t}}}$
	$\displaystyle =$	$\displaystyle \log\vert t\vert\mid_{2}^{3} = \log{3} - \log{2} = \log{3}{2}$

(a)

で

のとき $1 + x^{n}$ の評価を行なう．まず， $1 + x^{n} < 1 + x^{2}$ より

$\displaystyle \frac{1}{1+x^{2}} < \frac{1}{1+x^{n}}$

$\displaystyle \frac{1}{1+x^{n}} < 1$

$\displaystyle \int_{0}^{1}\frac{1}{1+x^{2}} dx < \int_{0}^{1}\frac{1}{1+x^n} dx < \int_{0}^{1} 1 dx$

$\displaystyle \int_{0}^{1}\frac{1}{1+x^2} dx = \tan^{-1}{x}\mid_{0}^{1} = \tan^{-1}{1} - \tan^{-1}{0} = \frac{\pi}{4}$

$\displaystyle \frac{\pi}{4} < \int_{0}^{1}\frac{1}{1+x^n} dx < 1$

$\displaystyle \frac{x^{n}}{2} < \frac{x^{n}}{1+x} < x^{n}$

$\displaystyle \int_{0}^{1}\frac{x^n}{2} dx < \int_{0}^{1}\frac{x^n}{1+x} dx < \int_{0}^{1} x^n dx$

$\displaystyle \int_{0}^{1}\frac{x^n}{2} dx = \frac{x^{n+1}}{2(n+1)}\mid_{0}^{1} = \frac{1}{2(n+1)}$

$\displaystyle \int_{0}^{1}x^{n} dx = \frac{x^{n+1}}{n+1}\mid_{0}^{1} = \frac{1}{n+1} < \frac{1}{n}$

$\displaystyle \frac{1}{2(n+1)} < \int_{0}^{1}\frac{x^{n}}{1+x} dx < \frac{1}{n+1} < \frac{1}{n}$

定積分と和の極限値の関係は次の式で与えられる．区間 $[a,b]$

上で定義された関数

の積分は，区間

を幅 $\frac{b-a}{n}$ の小区間で刻んでいくと，そこに生まれる長方形の面積(ただし $f(x) > 0$

)は

$\displaystyle f(a + \frac{b-a}{n})\frac{b-a}{n} + f(a + \frac{2(b-a)}{n})\frac{b-a}{n} + \cdots + = \frac{b-a}{n}\sum_{i=1}^{n}f(a + \frac{i(b-a)}{n})$

$\displaystyle \lim_{n \to \infty}\frac{b-a}{n}\sum_{i=1}^{n}f(a + \frac{i(b-a)}{n}) = \int_{a}^{b}f(x) dx$

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

$\displaystyle \lim_{n \to \infty}\frac{1}{n}(\frac{1}{1 + \frac{1}{n}} + \frac{1}{1 + \frac{2}{n}} + \cdots + \frac{1}{1 + \frac{n}{n}})$

$\displaystyle =$

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}f(1 + \frac{i}{n})$

$\displaystyle \lim_{n \to \infty}\frac{1}{n}(\frac{1}{1 + \frac{1}{n}} + \frac{1}{1 + \frac{2}{n}} + \cdots + \frac{1}{1 + \frac{n}{n}})$	$\displaystyle =$	$\displaystyle \int_{1}^{2}\frac{1}{x} dx = \log\vert x\vert\mid_{1}^{2}$
	$\displaystyle =$	$\displaystyle \log{2} - \log{1} = \log{2}$

別解 $f(x) = f(\frac{i}{n}) = \frac{1}{1 + \frac{i}{n}}$ より $f(x) = \frac{1}{1 + x}$ となるので

$\displaystyle \lim_{n \to \infty}\frac{1}{n}(\frac{1}{1 + \frac{1}{n}} + \frac{1}{1 + \frac{2}{n}} + \cdots + \frac{1}{1 + \frac{n}{n}})$	$\displaystyle =$	$\displaystyle \int_{1}^{2}\frac{1}{1+x} dx = \log\vert 1+x\vert\mid_{0}^{1}$
	$\displaystyle =$	$\displaystyle \log{2} - \log{1} = \log{2}$

$\displaystyle \sum_{i=1}^{n}\sqrt{\frac{1}{n^{2} + i^{2}}}$

$\displaystyle =$

$\displaystyle \frac{1}{n}\sum_{i=1}^{n}\sqrt{\frac{1}{1 + (\frac{i}{n})^{2}}}$

$\displaystyle \lim_{n \to \infty}\sum_{i=1}^{n}\sqrt{\frac{1}{n^{2} + i^{2}}}$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\sqrt{\frac{1}{1 + (\frac{i}{n})^{2}}}$
	$\displaystyle =$	$\displaystyle \int_{0}^{1}\sqrt{\frac{1}{1 + x^2}} dx$
	$\displaystyle =$	$\displaystyle \log\vert x + \sqrt{1 + x^2}\vert\mid_{0}^{1} = \log{1 + \sqrt{2}}$

$\displaystyle \lim_{x \rightarrow 0} \frac{\int_{0}^{x} \tan{(t^2)}dt}{x^{3}}$	$\displaystyle =$	$\displaystyle \lim_{x \to 0}\frac{\tan{x^2}}{3x^{2}}$
	$\displaystyle =$	$\displaystyle \lim_{x \to 0}\frac{2x \sec^{2}{2x}}{6x} = \lim_{x \to 0}\frac{2\sec^{2}{2x} + 8x\sec^{2}{2x}\tan{2x}}{6}$
	$\displaystyle =$	$\displaystyle \frac{1}{3}$