What is the integral from 0 to 1 of 2piy(y-y^9) ?

The integral from 0 to 1 of 2piy(y-y^9) is (16pi)/(33)

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integral from 0 to 1 of 2piy(y-y^9)

\int_{0}^{1} 2 π y (y - y^{9}) d y

\frac{16 π}{33}

Decimal

1.52319 \dots

Solution steps

\int_{0}^{1} 2 π y (y - y^{9}) 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}^{1} y (y - y^{9}) d y

Expand

y (y - y^{9}) : y^{2} - y^{10}

= 2 π \cdot \int_{0}^{1} y^{2} - y^{10} 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 (\int_{0}^{1} y^{2} d y - \int_{0}^{1} y^{10} d y)

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

\int_{0}^{1} y^{10} d y = \frac{1}{11}

= 2 π (\frac{1}{3} - \frac{1}{11})

Simplify

2 π (\frac{1}{3} - \frac{1}{11}) : \frac{16 π}{33}

= \frac{16 π}{33}

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What is the integral from 0 to 1 of 2piy(y-y^9) ?
The integral from 0 to 1 of 2piy(y-y^9) is (16pi)/(33)