integral from 0 to 2 of 3/4 x(2x-x^2)

解

\int_{0}^{2} \frac{3}{4} x (2 x - x^{2}) d x

1

解答ステップ

\int_{0}^{2} \frac{3}{4} x (2 x - x^{2}) d x

定数を除く:

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

= \frac{3}{4} \cdot \int_{0}^{2} x (2 x - x^{2}) d x

拡張

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

= \frac{3}{4} \cdot \int_{0}^{2} 2 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

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

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

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

= \frac{3}{4} (\frac{16}{3} - 4)

簡素化

\frac{3}{4} (\frac{16}{3} - 4) : 1

= 1