integral from 0 to x of (t^3-t)^5

解答

\int_{0}^{x} (t^{3} - t)^{5} d t

\frac{x ^{16}}{16} - \frac{5 x ^{14}}{14} + \frac{5 x ^{12}}{6} - x^{10} + \frac{5 x ^{8}}{8} - \frac{x ^{6}}{6}

求解步骤

\int_{0}^{x} (t^{3} - t)^{5} d t

乘开

(t^{3} - t)^{5} : t^{15} - 5 t^{13} + 10 t^{11} - 10 t^{9} + 5 t^{7} - t^{5}

= \int_{0}^{x} t^{15} - 5 t^{13} + 10 t^{11} - 10 t^{9} + 5 t^{7} - t^{5} d t

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x} t^{15} d t - \int_{0}^{x} 5 t^{13} d t + \int_{0}^{x} 10 t^{11} d t - \int_{0}^{x} 10 t^{9} d t + \int_{0}^{x} 5 t^{7} d t - \int_{0}^{x} t^{5} d t

\int_{0}^{x} t^{15} d t = \frac{x ^{16}}{16}

\int_{0}^{x} 5 t^{13} d t = \frac{5 x ^{14}}{14}

\int_{0}^{x} 10 t^{11} d t = \frac{5 x ^{12}}{6}

\int_{0}^{x} 10 t^{9} d t = x^{10}

\int_{0}^{x} 5 t^{7} d t = \frac{5 x ^{8}}{8}

\int_{0}^{x} t^{5} d t = \frac{x ^{6}}{6}

= \frac{x ^{16}}{16} - \frac{5 x ^{14}}{14} + \frac{5 x ^{12}}{6} - x^{10} + \frac{5 x ^{8}}{8} - \frac{x ^{6}}{6}

d er i v a t i v e 10 + \frac{50 x}{2 x ^{2} + 9}

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