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

The integral from 0 to 4 of 2pi(2-y)(4y^2) is -(512pi)/3

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

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

- \frac{512 π}{3}

Decimal Notation

- 536.16514 \dots

Solution steps

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

Simplify

= 8 π \cdot \int_{0}^{4} (2 - y) y^{2} d y

Expand

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

= 8 π \cdot \int_{0}^{4} 2 y^{2} - 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

= 8 π (\int_{0}^{4} 2 y^{2} d y - \int_{0}^{4} y^{3} d y)

\int_{0}^{4} 2 y^{2} d y = \frac{128}{3}

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

= 8 π (\frac{128}{3} - 64)

Simplify

8 π (\frac{128}{3} - 64) : - \frac{512 π}{3}

= - \frac{512 π}{3}

\int (25 - x^{2} - 3) d x

n = 3 \sum \infty (\frac{3}{11})^{- n}

\int \frac{8}{x ln ( 2 x )} d x

\frac{\partial}{\partial x} (e^{- x y})

x \to + 2 + lim (x^{2})

What is the integral from 0 to 4 of 2pi(2-y)(4y^2) ?
The integral from 0 to 4 of 2pi(2-y)(4y^2) is -(512pi)/3