What is the integral from 0 to 1 of 2pi(2-y)(7-7y^2) ?

The integral from 0 to 1 of 2pi(2-y)(7-7y^2) is (91pi)/6

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integral from 0 to 1 of 2pi(2-y)(7-7y^2)

\int_{0}^{1} 2 π (2 - y) (7 - 7 y^{2}) d y

\frac{91 π}{6}

Decimal Notation

47.64748 \dots

Solution steps

\int_{0}^{1} 2 π (2 - y) (7 - 7 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) (7 - 7 y^{2}) d y

Expand

(2 - y) (7 - 7 y^{2}) : 14 - 14 y^{2} - 7 y + 7 y^{3}

= 2 π \cdot \int_{0}^{1} 14 - 14 y^{2} - 7 y + 7 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} 14 d y - \int_{0}^{1} 14 y^{2} d y - \int_{0}^{1} 7 y d y + \int_{0}^{1} 7 y^{3} d y)

\int_{0}^{1} 14 d y = 14

\int_{0}^{1} 14 y^{2} d y = \frac{14}{3}

\int_{0}^{1} 7 y d y = \frac{7}{2}

\int_{0}^{1} 7 y^{3} d y = \frac{7}{4}

= 2 π (14 - \frac{14}{3} - \frac{7}{2} + \frac{7}{4})

Simplify

2 π (14 - \frac{14}{3} - \frac{7}{2} + \frac{7}{4}) : \frac{91 π}{6}

= \frac{91 π}{6}

\frac{3}{4 s}

\frac{d}{d x} (sec^{2} (x^{3}))

x \to 2 lim (- x + 7)

\int \frac{x ^{3}}{x ^{4} + 6} d x

f (x) = ln (x^{7}), (1, 0)

What is the integral from 0 to 1 of 2pi(2-y)(7-7y^2) ?
The integral from 0 to 1 of 2pi(2-y)(7-7y^2) is (91pi)/6