integral from 0 to 6 of x^3sqrt(x^2+36)

解

\int_{0}^{6} x^{3} x^{2} + 36 d x

\frac{31104 2 - 7776}{5} - 36 (1442 - 72)

十進法表記

2503.05662 \dots

解答ステップ

\int_{0}^{6} x^{3} x^{2} + 36 d x

u置換積分法を適用する

= \int_{6}^{62} (u^{2} - 36) u^{2} d u

拡張

(u^{2} - 36) u^{2} : u^{4} - 36 u^{2}

= \int_{6}^{62} u^{4} - 36 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_{6}^{62} u^{4} d u - \int_{6}^{62} 36 u^{2} d u

\int_{6}^{62} u^{4} d u = \frac{31104 2 - 7776}{5}

\int_{6}^{62} 36 u^{2} d u = 36 (1442 - 72)

= \frac{31104 2 - 7776}{5} - 36 (1442 - 72)