integral from 0 to 2 of t^3(1+t^4)^3

解

\int_{0}^{2} t^{3} (1 + t^{4})^{3} d t

5220

解答ステップ

\int_{0}^{2} t^{3} (1 + t^{4})^{3} d t

拡張

t^{3} (1 + t^{4})^{3} : t^{3} + 3 t^{7} + 3 t^{11} + t^{15}

= \int_{0}^{2} t^{3} + 3 t^{7} + 3 t^{11} + t^{15} d t

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_{0}^{2} t^{3} d t + \int_{0}^{2} 3 t^{7} d t + \int_{0}^{2} 3 t^{11} d t + \int_{0}^{2} t^{15} d t

\int_{0}^{2} t^{3} d t = 4

\int_{0}^{2} 3 t^{7} d t = 96

\int_{0}^{2} 3 t^{11} d t = 1024

\int_{0}^{2} t^{15} d t = 4096

= 4 + 96 + 1024 + 4096

簡素化

= 5220