integral of (3-sqrt(x))^2

解

\int (3 - x)^{2} d x

9 x - 4 x^{\frac{3}{2}} + \frac{x ^{2}}{2} + C

解答ステップ

\int (3 - x)^{2} d x

拡張

(3 - x)^{2} : 9 - 6 x + x

= \int 9 - 6 x + x 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 9 d x - \int 6 x d x + \int x d x

\int 9 d x = 9 x

\int 6 x d x = 4 x^{\frac{3}{2}}

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

= 9 x - 4 x^{\frac{3}{2}} + \frac{x ^{2}}{2}

定数を解答に追加する

= 9 x - 4 x^{\frac{3}{2}} + \frac{x ^{2}}{2} + C