【微分積分】4-5-1 広義積分の計算｜問題集
1.次の広義積分を求めなさい。
(1)\(\displaystyle \int_0^\infty e^{-x}dx\)
\(\displaystyle =\lim_{t\to\infty}\int_0^t e^{-x}dx\)
\(\displaystyle =\lim_{t\to\infty}(-e^{-t}+1)\)
\(\displaystyle =1\)
\(\displaystyle =\lim_{t\to\infty}(-e^{-t}+1)\)
\(\displaystyle =1\)
(2)\(\displaystyle \int_1^\infty\frac{1}{x}dx\)
\(\displaystyle =\lim_{t\to\infty}\int_1^t\frac{1}{x}dx\)
\(\displaystyle =\lim_{t\to\infty}\log t\)
\(\displaystyle =\infty\)
\(\displaystyle =\lim_{t\to\infty}\log t\)
\(\displaystyle =\infty\)
(3)\(\displaystyle \int_1^\infty\frac{1}{x^2}dx\)
\(\displaystyle =\lim_{t\to\infty}\int_1^t\frac{1}{x^2}dx\)
\(\displaystyle =\lim_{t\to\infty}\left(-\frac{1}{t}+1\right)\)
\(\displaystyle =1\)
\(\displaystyle =\lim_{t\to\infty}\left(-\frac{1}{t}+1\right)\)
\(\displaystyle =1\)
(4)\(\displaystyle \int_0^1\frac{1}{\sqrt{x}}dx\)
\(\displaystyle =\lim_{t\to+0}\int_t^1x^{-\frac{1}{2}}dx\)
\(\displaystyle =\lim_{t\to+0}(2-2t^{\frac{1}{2}})\)
\(\displaystyle =2\)
\(\displaystyle =\lim_{t\to+0}(2-2t^{\frac{1}{2}})\)
\(\displaystyle =2\)
(5)\(\displaystyle \int_0^1\frac{1}{x^2}dx\)
\(\displaystyle =\lim_{t\to+0}\int_t^1x^{-2}dx\)
\(\displaystyle =\lim_{t\to+0}\left(1-\frac{1}{t}\right)\)
\(\displaystyle =\infty\)
\(\displaystyle =\lim_{t\to+0}\left(1-\frac{1}{t}\right)\)
\(\displaystyle =\infty\)
次の学習に進もう！