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

The integral from 0 to 1 of 2pi(2-y)(2y^2) is (5pi)/3

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

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

\frac{5 π}{3}

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

Expand

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

= 4 π \cdot \int_{0}^{1} 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

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

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

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

= 4 π (\frac{2}{3} - \frac{1}{4})

Simplify

4 π (\frac{2}{3} - \frac{1}{4}) : \frac{5 π}{3}

= \frac{5 π}{3}

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\frac{d}{d x} (- 3 x^{6} ln (x))

\frac{d}{d x} ((4 - x^{\frac{2}{3}})^{\frac{3}{2}})

\frac{\partial}{\partial v} (\frac{1}{3} (v - 2 u))

What is the integral from 0 to 1 of 2pi(2-y)(2y^2) ?
The integral from 0 to 1 of 2pi(2-y)(2y^2) is (5pi)/3