integral from 0 to 4 of x^3sqrt(7-x^2)

解

\int_{0}^{4} x^{3} 7 - x^{2} d x

\frac{98 7}{15} + i \frac{558}{5}

解答ステップ

\int_{0}^{4} x^{3} 7 - x^{2} d x

u置換積分法を適用する

= \int_{7}^{3 i} - (- u^{2} + 7) u^{2} d u

定数を除く:

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

= - \int_{7}^{3 i} (- u^{2} + 7) u^{2} d u

拡張

(- u^{2} + 7) u^{2} : - u^{4} + 7 u^{2}

= - \int_{7}^{3 i} - u^{4} + 7 u^{2} 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_{7}^{3 i} u^{4} d u + \int_{7}^{3 i} 7 u^{2} d u)

\int_{7}^{3 i} u^{4} d u = - \frac{49 7}{5} + i \frac{243}{5}

\int_{7}^{3 i} 7 u^{2} d u = - \frac{49 7}{3} - 63 i

= - (- (- \frac{49 7}{5} + i \frac{243}{5}) - \frac{49 7}{3} - 63 i)

簡素化

- (- (- \frac{49 7}{5} + i \frac{243}{5}) - \frac{49 7}{3} - 63 i) : \frac{98 7}{15} + i \frac{558}{5}

= \frac{98 7}{15} + i \frac{558}{5}