What is the integral from 0 to 9 of 2piy(3-y/3) ?

The integral from 0 to 9 of 2piy(3-y/3) is 81pi

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

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

81 π

Decimal Notation

254.46900 \dots

Solution steps

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

Take the constant out:

\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

Take the constant out:

\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

Expand

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

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

Apply the Sum Rule:

\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)

Simplify

2 π 3 (\frac{81}{2} - 27) : 81 π

= 81 π

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What is the integral from 0 to 9 of 2piy(3-y/3) ?
The integral from 0 to 9 of 2piy(3-y/3) is 81pi