integral of (2-x)^5

解

\int (2 - x)^{5} d x

32 x - 40 x^{2} + \frac{80 x ^{3}}{3} - 10 x^{4} + 2 x^{5} - \frac{x ^{6}}{6} + C

解答ステップ

\int (2 - x)^{5} d x

拡張

(2 - x)^{5} : 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5}

= \int 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - 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 32 d x - \int 80 x d x + \int 80 x^{2} d x - \int 40 x^{3} d x + \int 10 x^{4} d x - \int x^{5} d x

\int 32 d x = 32 x

\int 80 x d x = 40 x^{2}

\int 80 x^{2} d x = \frac{80 x ^{3}}{3}

\int 40 x^{3} d x = 10 x^{4}

\int 10 x^{4} d x = 2 x^{5}

\int x^{5} d x = \frac{x ^{6}}{6}

= 32 x - 40 x^{2} + \frac{80 x ^{3}}{3} - 10 x^{4} + 2 x^{5} - \frac{x ^{6}}{6}

定数を解答に追加する

= 32 x - 40 x^{2} + \frac{80 x ^{3}}{3} - 10 x^{4} + 2 x^{5} - \frac{x ^{6}}{6} + C