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

解

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

\frac{57}{2}

十進法表記

28.5

解答ステップ

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

拡張

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

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

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

\int_{- 1}^{2} 4 x^{2} d x = 12

\int_{- 1}^{2} 4 x^{3} d x = 15

= \frac{3}{2} + 12 + 15

簡素化

\frac{3}{2} + 12 + 15 : \frac{57}{2}

= \frac{57}{2}