integral from 0 to 2 of kx(2-x)^2

解

\int_{0}^{2} k x (2 - x)^{2} d x

\frac{4}{3} k

解答ステップ

\int_{0}^{2} k x (2 - x)^{2} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= k \cdot \int_{0}^{2} x (2 - x)^{2} d x

拡張

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

= k \cdot \int_{0}^{2} 4 x - 4 x^{2} + 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

= k (\int_{0}^{2} 4 x d x - \int_{0}^{2} 4 x^{2} d x + \int_{0}^{2} x^{3} d x)

\int_{0}^{2} 4 x d x = 8

\int_{0}^{2} 4 x^{2} d x = \frac{32}{3}

\int_{0}^{2} x^{3} d x = 4

= k (8 - \frac{32}{3} + 4)

簡素化

k (8 - \frac{32}{3} + 4) : \frac{4}{3} k

= \frac{4}{3} k