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

解

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

x^{8} + x^{2} + \frac{x ^{26} + 1}{13} + \frac{6 ( x ^{14} + 1 )}{7} + \frac{2 ( x ^{20} - 1 )}{5}

解答ステップ

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

拡張

(1 + t^{3})^{4} : 1 + 4 t^{3} + 6 t^{6} + 4 t^{9} + t^{12}

= \int_{- 1}^{x^{2}} 1 + 4 t^{3} + 6 t^{6} + 4 t^{9} + t^{12} 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_{- 1}^{x^{2}} 1 d t + \int_{- 1}^{x^{2}} 4 t^{3} d t + \int_{- 1}^{x^{2}} 6 t^{6} d t + \int_{- 1}^{x^{2}} 4 t^{9} d t + \int_{- 1}^{x^{2}} t^{12} d t

\int_{- 1}^{x^{2}} 1 d t = x^{2} + 1

\int_{- 1}^{x^{2}} 4 t^{3} d t = x^{8} - 1

\int_{- 1}^{x^{2}} 6 t^{6} d t = \frac{6 ( x ^{14} + 1 )}{7}

\int_{- 1}^{x^{2}} 4 t^{9} d t = \frac{2 ( x ^{20} - 1 )}{5}

\int_{- 1}^{x^{2}} t^{12} d t = \frac{x ^{26} + 1}{13}

= x^{2} + 1 + x^{8} - 1 + \frac{6 ( x ^{14} + 1 )}{7} + \frac{2 ( x ^{20} - 1 )}{5} + \frac{x ^{26} + 1}{13}

簡素化

= x^{8} + x^{2} + \frac{x ^{26} + 1}{13} + \frac{6 ( x ^{14} + 1 )}{7} + \frac{2 ( x ^{20} - 1 )}{5}