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

解

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

\frac{16}{3}

十進法表記

5.33333 \dots

解答ステップ

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

拡張

x (x + 4)^{2} : x^{3} + 8 x^{2} + 16 x

= \int_{- 1}^{1} x^{3} + 8 x^{2} + 16 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}^{1} x^{3} d x + \int_{- 1}^{1} 8 x^{2} d x + \int_{- 1}^{1} 16 x d x

\int_{- 1}^{1} x^{3} d x = 0

\int_{- 1}^{1} 8 x^{2} d x = \frac{16}{3}

\int_{- 1}^{1} 16 x d x = 0

= 0 + \frac{16}{3} + 0

簡素化

= \frac{16}{3}