What is the integral from 0 to 1 of 2pi(2-y)(4-4y^2) ?

The integral from 0 to 1 of 2pi(2-y)(4-4y^2) is (26pi)/3

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integral from 0 to 1 of 2pi(2-y)(4-4y^2)

\int_{0}^{1} 2 π (2 - y) (4 - 4 y^{2}) d y

\frac{26 π}{3}

Decimal Notation

27.22713 \dots

Solution steps

\int_{0}^{1} 2 π (2 - y) (4 - 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}^{1} (2 - y) (4 - 4 y^{2}) d y

Expand

(2 - y) (4 - 4 y^{2}) : 8 - 8 y^{2} - 4 y + 4 y^{3}

= 2 π \cdot \int_{0}^{1} 8 - 8 y^{2} - 4 y + 4 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}^{1} 8 d y - \int_{0}^{1} 8 y^{2} d y - \int_{0}^{1} 4 y d y + \int_{0}^{1} 4 y^{3} d y)

\int_{0}^{1} 8 d y = 8

\int_{0}^{1} 8 y^{2} d y = \frac{8}{3}

\int_{0}^{1} 4 y d y = 2

\int_{0}^{1} 4 y^{3} d y = 1

= 2 π (8 - \frac{8}{3} - 2 + 1)

Simplify

2 π (8 - \frac{8}{3} - 2 + 1) : \frac{26 π}{3}

= \frac{26 π}{3}

y = - 2 x + 3, (0, 1)

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What is the integral from 0 to 1 of 2pi(2-y)(4-4y^2) ?
The integral from 0 to 1 of 2pi(2-y)(4-4y^2) is (26pi)/3