【高校数学Ⅲ】5-2-4 区分求積法｜問題集

\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}\)
\(\displaystyle f(x)=\frac{1}{\sqrt{1+x}}\)とおくと、
\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\)
\(\displaystyle =\int_0^1 \frac{1}{\sqrt{1+x}}dx\)
\(\displaystyle =[2\sqrt{1+x}]_0^1\)
\(=2\sqrt{2}-2\)

\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^4\)
\(\displaystyle f(x)=x^4\)とおくと、
\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\)
\(\displaystyle =\int_0^1 x^4dx\)
\(\displaystyle =\left[\frac{1}{5}x^5\right]_0^1\)
\(\displaystyle =\frac{1}{5}\)

\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{1}{1+\frac{k}{n}}\right)\)
\(\displaystyle f(x)=\frac{1}{1+x}\)とおくと、
\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\)
\(\displaystyle =\int_0^1 \frac{1}{1+x}dx\)
\(\displaystyle =[\log|1+x|]_0^1\)
\(\displaystyle =\log2\)

\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{2k}{n+\frac{k^2}{n}}\right)\)
\(\displaystyle f(x)=\frac{2x}{1+x^2}\)とおくと、
\(\displaystyle =\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\)
\(\displaystyle =\int_0^1 \frac{2x}{1+x^2}dx\)
\(\displaystyle =[\log|1+x^2|]_0^1\)
\(=\log2\)

\(0\leqq x\leqq1\)より
\(1\leqq x+1\leqq 2\)
\(x+1\leqq (x+1)^2\leqq 2(x+1)\)
よって、
\(x+1\leqq (x+1)^2\)

\(x+1\leqq (x+1)^2\)より
\(\displaystyle \frac{1}{x+1}\geqq \frac{1}{(x+1)^2}\)
\(\displaystyle \int_0^1\frac{1}{x+1}dx\geqq \int_0^1\frac{1}{(x+1)^2}dx\)
\(\displaystyle [\log|x+1|]_0^1\geqq [-\frac{1}{x+1}]_0^1\)
\(\displaystyle \log2\geqq\frac{1}{2}\)
よって、
\(\displaystyle \frac{1}{2}<\log2\)

\(x\geqq0\)より
\(x+1\leqq x^2+x+1\)
よって、
\(\displaystyle \frac{1}{x+1}\geqq \frac{1}{x^2+x+1}\)

\(\displaystyle \frac{1}{x+1}\geqq \frac{1}{x^2+x+1}\)より
\(\displaystyle \int_0^1 \frac{1}{x+1}dx\geqq \int_0^1 \frac{1}{x^2+x+1}dx\)
\(\displaystyle [\log|x+1|]_0^1\geqq \int_0^1 \frac{1}{x^2+x+1}dx\)
\(\displaystyle \log2\geqq\int_0^1 \frac{1}{x^2+x+1}dx\)
よって、
\(\displaystyle \log2>\int_0^1 \frac{dx}{x^2+x+1}\)

\(k\leqq x\leqq k+1\)とおくと、
\(\displaystyle \frac{1}{x}< \frac{1}{k}\)
\(\displaystyle \int_k^{k+1}\frac{1}{x}dx< \int_k^{k+1}\frac{1}{k}dx\)
\(\displaystyle [\log|x|]_1^{n+1}< 1+\frac{1}{2}+\cdots+\frac{1}{n}\)
\(\displaystyle \log(n+1)< 1+\frac{1}{2}+\cdots+\frac{1}{n}\)
よって、
\(\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}>\log(n+1)\)