integral of (1-x)(e-x+1)

解

\int (1 - x) (e - x + 1) d x

- \frac{e x ^{2}}{2} - x^{2} + \frac{x ^{3}}{3} + e x + x + C

解答ステップ

\int (1 - x) (e - x + 1) d x

拡張

(1 - x) (e - x + 1) : - e x - 2 x + x^{2} + e + 1

= \int - e x - 2 x + x^{2} + e + 1 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 e x d x - \int 2 x d x + \int x^{2} d x + \int e d x + \int 1 d x

\int e x d x = \frac{e x ^{2}}{2}

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

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

\int e d x = e x

\int 1 d x = x

= - \frac{e x ^{2}}{2} - x^{2} + \frac{x ^{3}}{3} + e x + x

定数を解答に追加する

= - \frac{e x ^{2}}{2} - x^{2} + \frac{x ^{3}}{3} + e x + x + C