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受欢迎的微积分 >

integral from 4 to 9 of x(2x^2+1)^6

解答

\int_{4}^{9} x (2 x^{2} + 1)^{6} d x

解答

\frac{16 3 ^{7} - 3 3 ^{7}}{28}

+1

十进制

1.09182 E 14

求解步骤

\int_{4}^{9} x (2 x^{2} + 1)^{6} d x

使用换元积分法

= \int_{16}^{81} \frac{( 2 u + 1 ) ^{6}}{2} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{16}^{81} (2 u + 1)^{6} d u

使用换元积分法

= \frac{1}{2} \cdot \int_{33}^{163} \frac{v ^{6}}{2} d v

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{33}^{163} v^{6} d v

使用幂法则

= \frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{7}}{7}]_{33}^{163}

化简

\frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{7}}{7}]_{33}^{163} : \frac{1}{4} [\frac{v ^{7}}{7}]_{33}^{163}

= \frac{1}{4} [\frac{v ^{7}}{7}]_{33}^{163}

带入上下限计算后得:

\frac{16 3 ^{7} - 3 3 ^{7}}{7}

= \frac{1}{4} \cdot \frac{16 3 ^{7} - 3 3 ^{7}}{7}

化简

= \frac{16 3 ^{7} - 3 3 ^{7}}{28}

作图

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