integral from 0 to 4 of 1/((x-2)^{2/3)}

解

\int_{0}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

6 \cdot 2^{\frac{1}{3}}

十進法表記

7.55952 \dots

解答ステップ

\int_{0}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

未定義の (特異) 点を求める:

x = 2

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}^{2} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x + \int_{2}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

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

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

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

簡素化

- 3 (- 1)^{\frac{1}{3}} \cdot 2^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}} : 6 \cdot 2^{\frac{1}{3}}

= 6 \cdot 2^{\frac{1}{3}}