What is the integral from 0 to 5 of 2piy(25-y^2) ?

The integral from 0 to 5 of 2piy(25-y^2) is (625pi)/2

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integral from 0 to 5 of 2piy(25-y^2)

\int_{0}^{5} 2 π y (25 - y^{2}) d y

\frac{625 π}{2}

Decimal Notation

981.74770 \dots

Solution steps

\int_{0}^{5} 2 π y (25 - 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}^{5} y (25 - y^{2}) d y

Expand

y (25 - y^{2}) : 25 y - y^{3}

= 2 π \cdot \int_{0}^{5} 25 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}^{5} 25 y d y - \int_{0}^{5} y^{3} d y)

\int_{0}^{5} 25 y d y = \frac{625}{2}

\int_{0}^{5} y^{3} d y = \frac{625}{4}

= 2 π (\frac{625}{2} - \frac{625}{4})

Simplify

2 π (\frac{625}{2} - \frac{625}{4}) : \frac{625 π}{2}

= \frac{625 π}{2}

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What is the integral from 0 to 5 of 2piy(25-y^2) ?
The integral from 0 to 5 of 2piy(25-y^2) is (625pi)/2