What is the integral from 0 to 2 of 2piy(4-y^2) ?

The integral from 0 to 2 of 2piy(4-y^2) is 8pi

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integral from 0 to 2 of 2piy(4-y^2)

\int_{0}^{2} 2 π y (4 - y^{2}) d y

8 π

Decimal

25.13274 \dots

Solution steps

\int_{0}^{2} 2 π y (4 - 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}^{2} y (4 - y^{2}) d y

Expand

y (4 - y^{2}) : 4 y - y^{3}

= 2 π \cdot \int_{0}^{2} 4 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}^{2} 4 y d y - \int_{0}^{2} y^{3} d y)

\int_{0}^{2} 4 y d y = 8

\int_{0}^{2} y^{3} d y = 4

= 2 π (8 - 4)

Simplify

= 8 π

\frac{d}{d x} (\frac{x + 1}{x - 2})

\int \frac{3}{x ^{6}} d x

y = x^{2} - 2 x - 2, (3, 1)

\int \frac{1}{2} (3 sin (x) + 7) d x

y = 2 x (5 - 3 x)^{4}

What is the integral from 0 to 2 of 2piy(4-y^2) ?
The integral from 0 to 2 of 2piy(4-y^2) is 8pi