【微分積分】4-5-3 広義の置換積分｜問題集

\(\displaystyle =\lim_{t\to1-}\int_0^t\frac{1}{\sqrt{1-x}}dx\)
\(u=\sqrt{1-x}\)とおくと、\(\displaystyle dx=-2udu\)
\(\displaystyle =\lim_{t\to1-}\int_1^{\sqrt{1-t}}-\frac{2u}{u}du\)
\(\displaystyle =\lim_{t\to1-}[-2u]_1^{\sqrt{1-t}}\)
\(\displaystyle =-2(0-1)\)
\(\displaystyle =2\)

\(\displaystyle =\lim_{t\to0+}\int_t^1\frac{1}{x}dx\)
\(\displaystyle =\lim_{t\to0+}[\log x]_t^1\)
\(\displaystyle =\log1-\lim_{t\to0+}\log|t|\)
\(\displaystyle =\infty\)

\(\displaystyle =\lim_{t\to0+}\int_t^1\log xdx\)
\(\displaystyle =\lim_{t\to0+}\left([x\log x]_t^1-\int_t^1dx\right)\)
\(\displaystyle =\lim_{t\to0+}\{\log1-1-(t\log t-t)\}\)
\(\displaystyle =0-1-(0-0)\)
\(\displaystyle =-1\)