integral from 0 to 1 of (x+2)(x+1)^9

解

\int_{0}^{1} (x + 2) (x + 1)^{9} d x

\frac{31723}{110}

十進法表記

288.39090 \dots

解答ステップ

\int_{0}^{1} (x + 2) (x + 1)^{9} d x

拡張

(x + 2) (x + 1)^{9} : x^{10} + 11 x^{9} + 54 x^{8} + 156 x^{7} + 294 x^{6} + 378 x^{5} + 336 x^{4} + 204 x^{3} + 81 x^{2} + 19 x + 2

= \int_{0}^{1} x^{10} + 11 x^{9} + 54 x^{8} + 156 x^{7} + 294 x^{6} + 378 x^{5} + 336 x^{4} + 204 x^{3} + 81 x^{2} + 19 x + 2 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^{10} d x + \int_{0}^{1} 11 x^{9} d x + \int_{0}^{1} 54 x^{8} d x + \int_{0}^{1} 156 x^{7} d x + \int_{0}^{1} 294 x^{6} d x + \int_{0}^{1} 378 x^{5} d x + \int_{0}^{1} 336 x^{4} d x + \int_{0}^{1} 204 x^{3} d x + \int_{0}^{1} 81 x^{2} d x + \int_{0}^{1} 19 x d x + \int_{0}^{1} 2 d x

\int_{0}^{1} x^{10} d x = \frac{1}{11}

\int_{0}^{1} 11 x^{9} d x = \frac{11}{10}

\int_{0}^{1} 54 x^{8} d x = 6

\int_{0}^{1} 156 x^{7} d x = \frac{39}{2}

\int_{0}^{1} 294 x^{6} d x = 42

\int_{0}^{1} 378 x^{5} d x = 63

\int_{0}^{1} 336 x^{4} d x = \frac{336}{5}

\int_{0}^{1} 204 x^{3} d x = 51

\int_{0}^{1} 81 x^{2} d x = 27

\int_{0}^{1} 19 x d x = \frac{19}{2}

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

= \frac{1}{11} + \frac{11}{10} + 6 + \frac{39}{2} + 42 + 63 + \frac{336}{5} + 51 + 27 + \frac{19}{2} + 2

簡素化

\frac{1}{11} + \frac{11}{10} + 6 + \frac{39}{2} + 42 + 63 + \frac{336}{5} + 51 + 27 + \frac{19}{2} + 2 : \frac{31723}{110}

= \frac{31723}{110}