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

The integral from 0 to 1 of 2pi(2-y)(5-5y^2) is (65pi)/6

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integral from 0 to 1 of 2pi(2-y)(5-5y^2)

\int_{0}^{1} 2 π (2 - y) (5 - 5 y^{2}) d y

\frac{65 π}{6}

Decimal Notation

34.03392 \dots

Solution steps

\int_{0}^{1} 2 π (2 - y) (5 - 5 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) (5 - 5 y^{2}) d y

Expand

(2 - y) (5 - 5 y^{2}) : 10 - 10 y^{2} - 5 y + 5 y^{3}

= 2 π \cdot \int_{0}^{1} 10 - 10 y^{2} - 5 y + 5 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} 10 d y - \int_{0}^{1} 10 y^{2} d y - \int_{0}^{1} 5 y d y + \int_{0}^{1} 5 y^{3} d y)

\int_{0}^{1} 10 d y = 10

\int_{0}^{1} 10 y^{2} d y = \frac{10}{3}

\int_{0}^{1} 5 y d y = \frac{5}{2}

\int_{0}^{1} 5 y^{3} d y = \frac{5}{4}

= 2 π (10 - \frac{10}{3} - \frac{5}{2} + \frac{5}{4})

Simplify

2 π (10 - \frac{10}{3} - \frac{5}{2} + \frac{5}{4}) : \frac{65 π}{6}

= \frac{65 π}{6}

\int_{0}^{4} 0.125 x^{2} d x

y^{^{''}} + 4 y = sin (x)

\frac{2 x ^{2} + 1}{x ^{2} + 8}

\int_{0}^{1} ln (x^{2} + 1) d x

\int \frac{1}{x ^{2} + x + 1} d x

What is the integral from 0 to 1 of 2pi(2-y)(5-5y^2) ?
The integral from 0 to 1 of 2pi(2-y)(5-5y^2) is (65pi)/6