integral from 0 to 4 of 1/((9-3x)^{1/3)}

解

\int_{0}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

\frac{9 ^{\frac{2}{3}} - 3 ^{\frac{2}{3}}}{2}

十進法表記

1.12333 \dots

解答ステップ

\int_{0}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

未定義の (特異) 点を求める:

x = 3

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}^{3} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x + \int_{3}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

\int_{0}^{3} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x = \frac{9 ^{\frac{2}{3}}}{2}

\int_{3}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x = - \frac{( - 1 ) ^{\frac{2}{3}} \cdot 3 ^{\frac{2}{3}}}{2}

= \frac{9 ^{\frac{2}{3}}}{2} - \frac{( - 1 ) ^{\frac{2}{3}} \cdot 3 ^{\frac{2}{3}}}{2}

簡素化

\frac{9 ^{\frac{2}{3}}}{2} - \frac{( - 1 ) ^{\frac{2}{3}} \cdot 3 ^{\frac{2}{3}}}{2} : \frac{9 ^{\frac{2}{3}} - 3 ^{\frac{2}{3}}}{2}

= \frac{9 ^{\frac{2}{3}} - 3 ^{\frac{2}{3}}}{2}