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

解答

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

\frac{6063}{5}

十进制

1212.6

求解步骤

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

提出常数:

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

= 2 \cdot \int_{1}^{2} x (4 x + 1)^{3} d x

乘开

x (4 x + 1)^{3} : 64 x^{4} + 48 x^{3} + 12 x^{2} + x

= 2 \cdot \int_{1}^{2} 64 x^{4} + 48 x^{3} + 12 x^{2} + x d x

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

= 2 (\int_{1}^{2} 64 x^{4} d x + \int_{1}^{2} 48 x^{3} d x + \int_{1}^{2} 12 x^{2} d x + \int_{1}^{2} x d x)

\int_{1}^{2} 64 x^{4} d x = \frac{1984}{5}

\int_{1}^{2} 48 x^{3} d x = 180

\int_{1}^{2} 12 x^{2} d x = 28

\int_{1}^{2} x d x = \frac{3}{2}

= 2 (\frac{1984}{5} + 180 + 28 + \frac{3}{2})

化简

2 (\frac{1984}{5} + 180 + 28 + \frac{3}{2}) : \frac{6063}{5}

= \frac{6063}{5}

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