integral from 0 to 4 of 2pix(4x-x^2)

解答

\int_{0}^{4} 2 π x (4 x - x^{2}) d x

\frac{128 π}{3}

十进制

134.04128 \dots

求解步骤

\int_{0}^{4} 2 π x (4 x - x^{2}) d x

提出常数:

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

= 2 π \cdot \int_{0}^{4} x (4 x - x^{2}) d x

乘开

x (4 x - x^{2}) : 4 x^{2} - x^{3}

= 2 π \cdot \int_{0}^{4} 4 x^{2} - x^{3} 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 π (\int_{0}^{4} 4 x^{2} d x - \int_{0}^{4} x^{3} d x)

\int_{0}^{4} 4 x^{2} d x = \frac{256}{3}

\int_{0}^{4} x^{3} d x = 64

= 2 π (\frac{256}{3} - 64)

化简

2 π (\frac{256}{3} - 64) : \frac{128 π}{3}

= \frac{128 π}{3}

s l o p e y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (3, - 2)

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