integral from 0 to 4 of 1/(\sqrt[3]{x-1)}

解

\int_{0}^{4} \frac{1}{3 x - 1} d x

- \frac{3}{2} + \frac{3 \cdot 3 ^{\frac{2}{3}}}{2}

十進法表記

1.62012 \dots

解答ステップ

\int_{0}^{4} \frac{1}{3 x - 1} d x

未定義の (特異) 点を求める:

x = 1

存在する場合は

b, a < b < c, f (b) =

未定義

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{1} \frac{1}{3 x - 1} d x + \int_{1}^{4} \frac{1}{3 x - 1} d x

\int_{0}^{1} \frac{1}{3 x - 1} d x = - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2}

\int_{1}^{4} \frac{1}{3 x - 1} d x = \frac{3 \cdot 3 ^{\frac{2}{3}}}{2}

= - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + \frac{3 \cdot 3 ^{\frac{2}{3}}}{2}

3 (- 1)^{\frac{2}{3}} = 3

= - \frac{3}{2} + \frac{3 \cdot 3 ^{\frac{2}{3}}}{2}