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

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

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

\int_{0}^{3} 2 π y (9 - y^{2}) d y

\frac{81 π}{2}

Decimal Notation

127.23450 \dots

Solution steps

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

Expand

y (9 - y^{2}) : 9 y - y^{3}

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

\int_{0}^{3} 9 y d y = \frac{81}{2}

\int_{0}^{3} y^{3} d y = \frac{81}{4}

= 2 π (\frac{81}{2} - \frac{81}{4})

Simplify

2 π (\frac{81}{2} - \frac{81}{4}) : \frac{81 π}{2}

= \frac{81 π}{2}

(2, - 2), (9, 3)

y^{(\frac{1}{2})} y^{^{'}} + y^{(\frac{3}{2})} = 1

4 arcsin (x^{2})

\int_{0}^{6} \frac{1}{x ^{2} - 6 x + 5} d x

\frac{d}{d x} (A t e^{- t})

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