integral from 4 to 8 of t^2ln(3t)

解答

\int_{4}^{8} t^{2} ln (3 t) d t

- \frac{448}{9} + \frac{512 ln ( 24 ) - 64 ln ( 12 )}{3}

十进制

439.59873 \dots

求解步骤

\int_{4}^{8} t^{2} ln (3 t) d t

使用分布积分法

= [\frac{1}{3} t^{3} ln (3 t) - \int \frac{t ^{2}}{3} d t]_{4}^{8}

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

= [\frac{1}{3} t^{3} ln (3 t) - \frac{t ^{3}}{9}]_{4}^{8}

带入上下限计算后得:

- \frac{448}{9} + \frac{512 ln ( 24 ) - 64 ln ( 12 )}{3}

= - \frac{448}{9} + \frac{512 ln ( 24 ) - 64 ln ( 12 )}{3}

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