integral from 0 to 1 of 2pi(2-y)(7y^2)

解答

\int_{0}^{1} 2 π (2 - y) (7 y^{2}) d y

\frac{35 π}{6}

十进制

18.32595 \dots

求解步骤

\int_{0}^{1} 2 π (2 - y) \cdot 7 y^{2} d y

提出常数:

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

= 2 π 7 \cdot \int_{0}^{1} (2 - y) y^{2} d y

化简

= 14 π \cdot \int_{0}^{1} (2 - y) y^{2} d y

乘开

(2 - y) y^{2} : 2 y^{2} - y^{3}

= 14 π \cdot \int_{0}^{1} 2 y^{2} - y^{3} d y

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

= 14 π (\int_{0}^{1} 2 y^{2} d y - \int_{0}^{1} y^{3} d y)

\int_{0}^{1} 2 y^{2} d y = \frac{2}{3}

\int_{0}^{1} y^{3} d y = \frac{1}{4}

= 14 π (\frac{2}{3} - \frac{1}{4})

化简

14 π (\frac{2}{3} - \frac{1}{4}) : \frac{35 π}{6}

= \frac{35 π}{6}