integral from 1 to 2 of x(x-1)^4

解

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

\frac{11}{30}

十進法表記

0.36666 \dots

解答ステップ

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

拡張

x (x - 1)^{4} : x^{5} - 4 x^{4} + 6 x^{3} - 4 x^{2} + x

= \int_{1}^{2} x^{5} - 4 x^{4} + 6 x^{3} - 4 x^{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_{1}^{2} x^{5} d x - \int_{1}^{2} 4 x^{4} d x + \int_{1}^{2} 6 x^{3} d x - \int_{1}^{2} 4 x^{2} d x + \int_{1}^{2} x d x

\int_{1}^{2} x^{5} d x = \frac{21}{2}

\int_{1}^{2} 4 x^{4} d x = \frac{124}{5}

\int_{1}^{2} 6 x^{3} d x = \frac{45}{2}

\int_{1}^{2} 4 x^{2} d x = \frac{28}{3}

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

= \frac{21}{2} - \frac{124}{5} + \frac{45}{2} - \frac{28}{3} + \frac{3}{2}

簡素化

\frac{21}{2} - \frac{124}{5} + \frac{45}{2} - \frac{28}{3} + \frac{3}{2} : \frac{11}{30}

= \frac{11}{30}