integral from 0 to 1 of (1+sqrt(x))^8

解

\int_{0}^{1} (1 + x)^{8} d x

\frac{4097}{45}

十進法表記

91.04444 \dots

解答ステップ

\int_{0}^{1} (1 + x)^{8} d x

拡張

(1 + x)^{8} : 1 + 8 x + 28 x + 56 x^{\frac{3}{2}} + 70 x^{2} + 56 x^{\frac{5}{2}} + 28 x^{3} + 8 x^{\frac{7}{2}} + x^{4}

= \int_{0}^{1} 1 + 8 x + 28 x + 56 x^{\frac{3}{2}} + 70 x^{2} + 56 x^{\frac{5}{2}} + 28 x^{3} + 8 x^{\frac{7}{2}} + x^{4} 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} 1 d x + \int_{0}^{1} 8 x d x + \int_{0}^{1} 28 x d x + \int_{0}^{1} 56 x^{\frac{3}{2}} d x + \int_{0}^{1} 70 x^{2} d x + \int_{0}^{1} 56 x^{\frac{5}{2}} d x + \int_{0}^{1} 28 x^{3} d x + \int_{0}^{1} 8 x^{\frac{7}{2}} d x + \int_{0}^{1} x^{4} d x

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 8 x d x = \frac{16}{3}

\int_{0}^{1} 28 x d x = 14

\int_{0}^{1} 56 x^{\frac{3}{2}} d x = \frac{112}{5}

\int_{0}^{1} 70 x^{2} d x = \frac{70}{3}

\int_{0}^{1} 56 x^{\frac{5}{2}} d x = 16

\int_{0}^{1} 28 x^{3} d x = 7

\int_{0}^{1} 8 x^{\frac{7}{2}} d x = \frac{16}{9}

\int_{0}^{1} x^{4} d x = \frac{1}{5}

= 1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5}

簡素化

1 + \frac{16}{3} + 14 + \frac{112}{5} + \frac{70}{3} + 16 + 7 + \frac{16}{9} + \frac{1}{5} : \frac{4097}{45}

= \frac{4097}{45}