integral of-1/4 (x-1)(x-2)(x-4)

解

\int - \frac{1}{4} (x - 1) (x - 2) (x - 4) d x

- \frac{1}{4} (\frac{x ^{4}}{4} - \frac{7 x ^{3}}{3} + 7 x^{2} - 8 x) + C

解答ステップ

\int - \frac{1}{4} (x - 1) (x - 2) (x - 4) d x

定数を除く:

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

= - \frac{1}{4} \cdot \int (x - 1) (x - 2) (x - 4) d x

拡張

(x - 1) (x - 2) (x - 4) : x^{3} - 7 x^{2} + 14 x - 8

= - \frac{1}{4} \cdot \int x^{3} - 7 x^{2} + 14 x - 8 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{1}{4} (\int x^{3} d x - \int 7 x^{2} d x + \int 14 x d x - \int 8 d x)

\int x^{3} d x = \frac{x ^{4}}{4}

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 14 x d x = 7 x^{2}

\int 8 d x = 8 x

= - \frac{1}{4} (\frac{x ^{4}}{4} - \frac{7 x ^{3}}{3} + 7 x^{2} - 8 x)

定数を解答に追加する

= - \frac{1}{4} (\frac{x ^{4}}{4} - \frac{7 x ^{3}}{3} + 7 x^{2} - 8 x) + C