What is the integral from 0 to 4 of 2pi(y)(3y+4-y^2) ?

The integral from 0 to 4 of 2pi(y)(3y+4-y^2) is 64pi

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integral from 0 to 4 of 2pi(y)(3y+4-y^2)

\int_{0}^{4} 2 π (y) (3 y + 4 - y^{2}) d y

64 π

Decimal Notation

201.06192 \dots

Solution steps

\int_{0}^{4} 2 π y (3 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}^{4} y (3 y + 4 - y^{2}) d y

Expand

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

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

\int_{0}^{4} 3 y^{2} d y = 64

\int_{0}^{4} 4 y d y = 32

\int_{0}^{4} y^{3} d y = 64

= 2 π (64 + 32 - 64)

Simplify

= 64 π

\int_{0}^{\frac{1}{2}} x cos (π) x d x

\int_{0}^{1.75} 600 x d x

x \to 0 lim (e^{\frac{8}{x}})

f (y) = \frac{A}{y ^{9}} + B e^{y}

x \to 2 - lim (\frac{3}{x - 2})

What is the integral from 0 to 4 of 2pi(y)(3y+4-y^2) ?
The integral from 0 to 4 of 2pi(y)(3y+4-y^2) is 64pi