integral from 0 to 9 of 2piy(3-y/3)

解答

\int_{0}^{9} 2 π y (3 - \frac{y}{3}) d y

81 π

十进制

254.46900 \dots

求解步骤

\int_{0}^{9} 2 π y (3 - \frac{y}{3}) d y

提出常数:

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

= 2 π \cdot \int_{0}^{9} y (3 - \frac{y}{3}) d y

3 - \frac{y}{3} = 3 (1 - \frac{\frac{y}{3}}{3})

= 2 π \cdot \int_{0}^{9} y \cdot 3 (1 - \frac{\frac{y}{3}}{3}) d y

提出常数:

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

= 2 π 3 \cdot \int_{0}^{9} y (1 - \frac{\frac{y}{3}}{3}) d y

\frac{\frac{y}{3}}{3} = \frac{y}{9}

= 2 π 3 \cdot \int_{0}^{9} y (1 - \frac{y}{9}) d y

乘开

y (1 - \frac{y}{9}) : y - \frac{y ^{2}}{9}

= 2 π 3 \cdot \int_{0}^{9} y - \frac{y ^{2}}{9} 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

= 2 π 3 (\int_{0}^{9} y d y - \int_{0}^{9} \frac{y ^{2}}{9} d y)

\int_{0}^{9} y d y = \frac{81}{2}

\int_{0}^{9} \frac{y ^{2}}{9} d y = 27

= 2 π 3 (\frac{81}{2} - 27)

化简

2 π 3 (\frac{81}{2} - 27) : 81 π

= 81 π

\frac{d}{d x} (- \frac{5 e ^{\frac{5}{x}}}{x ^{2}})

\frac{\partial}{\partial x} (30 - (x^{2} + y^{2}))

\int sin^{5} (11 x) d x

\int ([sin (x)]^{2} x + [cos (x)]^{2} x) d x

\int \frac{( 1 + x ) ^{3}}{x} d x