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

解答

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

\frac{92}{945}

十进制

0.09735 \dots

求解步骤

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

乘开

(2 t^{3} - \frac{4}{3} t)^{2} : 4 t^{6} - \frac{16 t ^{4}}{3} + \frac{16 t ^{2}}{9}

= \int_{0}^{1} 4 t^{6} - \frac{16 t ^{4}}{3} + \frac{16 t ^{2}}{9} 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}^{1} 4 t^{6} d t - \int_{0}^{1} \frac{16 t ^{4}}{3} d t + \int_{0}^{1} \frac{16 t ^{2}}{9} d t

\int_{0}^{1} 4 t^{6} d t = \frac{4}{7}

\int_{0}^{1} \frac{16 t ^{4}}{3} d t = \frac{16}{15}

\int_{0}^{1} \frac{16 t ^{2}}{9} d t = \frac{16}{27}

= \frac{4}{7} - \frac{16}{15} + \frac{16}{27}

化简

\frac{4}{7} - \frac{16}{15} + \frac{16}{27} : \frac{92}{945}

= \frac{92}{945}

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