integral of (4-4x)^5

解

\int (4 - 4 x)^{5} d x

1024 x - 2560 x^{2} + \frac{10240 x ^{3}}{3} - 2560 x^{4} + 1024 x^{5} - \frac{512 x ^{6}}{3} + C

解答ステップ

\int (4 - 4 x)^{5} d x

拡張

(4 - 4 x)^{5} : 1024 - 5120 x + 10240 x^{2} - 10240 x^{3} + 5120 x^{4} - 1024 x^{5}

= \int 1024 - 5120 x + 10240 x^{2} - 10240 x^{3} + 5120 x^{4} - 1024 x^{5} 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 1024 d x - \int 5120 x d x + \int 10240 x^{2} d x - \int 10240 x^{3} d x + \int 5120 x^{4} d x - \int 1024 x^{5} d x

\int 1024 d x = 1024 x

\int 5120 x d x = 2560 x^{2}

\int 10240 x^{2} d x = \frac{10240 x ^{3}}{3}

\int 10240 x^{3} d x = 2560 x^{4}

\int 5120 x^{4} d x = 1024 x^{5}

\int 1024 x^{5} d x = \frac{512 x ^{6}}{3}

= 1024 x - 2560 x^{2} + \frac{10240 x ^{3}}{3} - 2560 x^{4} + 1024 x^{5} - \frac{512 x ^{6}}{3}

定数を解答に追加する

= 1024 x - 2560 x^{2} + \frac{10240 x ^{3}}{3} - 2560 x^{4} + 1024 x^{5} - \frac{512 x ^{6}}{3} + C