integral of (2-2x)^7

解

\int (2 - 2 x)^{7} d x

128 x - 448 x^{2} + 896 x^{3} - 1120 x^{4} + 896 x^{5} - 448 x^{6} + 128 x^{7} - 16 x^{8} + C

解答ステップ

\int (2 - 2 x)^{7} d x

拡張

(2 - 2 x)^{7} : 128 - 896 x + 2688 x^{2} - 4480 x^{3} + 4480 x^{4} - 2688 x^{5} + 896 x^{6} - 128 x^{7}

= \int 128 - 896 x + 2688 x^{2} - 4480 x^{3} + 4480 x^{4} - 2688 x^{5} + 896 x^{6} - 128 x^{7} 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 128 d x - \int 896 x d x + \int 2688 x^{2} d x - \int 4480 x^{3} d x + \int 4480 x^{4} d x - \int 2688 x^{5} d x + \int 896 x^{6} d x - \int 128 x^{7} d x

\int 128 d x = 128 x

\int 896 x d x = 448 x^{2}

\int 2688 x^{2} d x = 896 x^{3}

\int 4480 x^{3} d x = 1120 x^{4}

\int 4480 x^{4} d x = 896 x^{5}

\int 2688 x^{5} d x = 448 x^{6}

\int 896 x^{6} d x = 128 x^{7}

\int 128 x^{7} d x = 16 x^{8}

= 128 x - 448 x^{2} + 896 x^{3} - 1120 x^{4} + 896 x^{5} - 448 x^{6} + 128 x^{7} - 16 x^{8}

定数を解答に追加する

= 128 x - 448 x^{2} + 896 x^{3} - 1120 x^{4} + 896 x^{5} - 448 x^{6} + 128 x^{7} - 16 x^{8} + C