integral from 0 to 1 of x^2sqrt(x-1)

解

\int_{0}^{1} x^{2} x - 1 d x

\frac{16 i}{105}

解答ステップ

\int_{0}^{1} x^{2} x - 1 d x

u置換積分法を適用する

= \int_{- 1}^{0} (u + 1)^{2} u d u

拡張

(u + 1)^{2} u : u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u

= \int_{- 1}^{0} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}} + u d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{0} u^{\frac{5}{2}} d u + \int_{- 1}^{0} 2 u^{\frac{3}{2}} d u + \int_{- 1}^{0} u d u

\int_{- 1}^{0} u^{\frac{5}{2}} d u = \frac{2 i}{7}

\int_{- 1}^{0} 2 u^{\frac{3}{2}} d u = - \frac{4 i}{5}

\int_{- 1}^{0} u d u = \frac{2 i}{3}

= \frac{2 i}{7} - \frac{4 i}{5} + \frac{2 i}{3}

簡素化

\frac{2 i}{7} - \frac{4 i}{5} + \frac{2 i}{3} : \frac{16 i}{105}

= \frac{16 i}{105}