integral from 0 to 1 of (sqrt(x)(x+1))

解

\int_{0}^{1} (x (x + 1)) d x

\frac{16}{15}

十進法表記

1.06666 \dots

解答ステップ

\int_{0}^{1} x (x + 1) d x

拡張

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

= \int_{0}^{1} x^{\frac{3}{2}} + x d x

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_{0}^{1} x^{\frac{3}{2}} d x + \int_{0}^{1} x d x

\int_{0}^{1} x^{\frac{3}{2}} d x = \frac{2}{5}

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

= \frac{2}{5} + \frac{2}{3}

簡素化

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}