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

The integral from 0 to 1 of 2pi(2-y)(7y^2) is (35pi)/6

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

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

\frac{35 π}{6}

Decimal Notation

18.32595 \dots

Solution steps

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

Simplify

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

Expand

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

= 14 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

= 14 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}

= 14 π (\frac{2}{3} - \frac{1}{4})

Simplify

14 π (\frac{2}{3} - \frac{1}{4}) : \frac{35 π}{6}

= \frac{35 π}{6}

x = 49 - y^{2}, x = y^{2} - 49

x \to 0 lim (lo g_{5} (3 x))

\frac{d}{d x} (e^{x^{2} - 4} + 5 x + 6)

\int \frac{cos ( x )}{7 + sin ( x )} d x

What is the integral from 0 to 1 of 2pi(2-y)(7y^2) ?
The integral from 0 to 1 of 2pi(2-y)(7y^2) is (35pi)/6