integral from 1 to 4 of xsqrt(3-x)

解

\int_{1}^{4} x 3 - x d x

\frac{12 2}{5} + i \frac{12}{5}

解答ステップ

\int_{1}^{4} x 3 - x d x

u置換積分法を適用する

= \int_{2}^{- 1} - (- u + 3) u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{2} - (- u + 3) u d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{- 1}^{2} (- u + 3) u d u)

拡張

(- u + 3) u : - u^{\frac{3}{2}} + 3 u

= - (- \int_{- 1}^{2} - u^{\frac{3}{2}} + 3 u d u)

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_{- 1}^{2} u^{\frac{3}{2}} d u + \int_{- 1}^{2} 3 u d u))

\int_{- 1}^{2} u^{\frac{3}{2}} d u = \frac{8 2}{5} - i \frac{2}{5}

\int_{- 1}^{2} 3 u d u = 42 + 2 i

= - (- (- (\frac{8 2}{5} - i \frac{2}{5}) + 42 + 2 i))

簡素化

- (- (- (\frac{8 2}{5} - i \frac{2}{5}) + 42 + 2 i)) : \frac{12 2}{5} + i \frac{12}{5}

= \frac{12 2}{5} + i \frac{12}{5}