What is the integral from 0 to 1 of 2pi(2-y)(8-8y^2) ?

The integral from 0 to 1 of 2pi(2-y)(8-8y^2) is (52pi)/3

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integral from 0 to 1 of 2pi(2-y)(8-8y^2)

\int_{0}^{1} 2 π (2 - y) (8 - 8 y^{2}) d y

\frac{52 π}{3}

Decimal

54.45427 \dots

Solution steps

\int_{0}^{1} 2 π (2 - y) (8 - 8 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) (8 - 8 y^{2}) d y

Expand

(2 - y) (8 - 8 y^{2}) : 16 - 16 y^{2} - 8 y + 8 y^{3}

= 2 π \cdot \int_{0}^{1} 16 - 16 y^{2} - 8 y + 8 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} 16 d y - \int_{0}^{1} 16 y^{2} d y - \int_{0}^{1} 8 y d y + \int_{0}^{1} 8 y^{3} d y)

\int_{0}^{1} 16 d y = 16

\int_{0}^{1} 16 y^{2} d y = \frac{16}{3}

\int_{0}^{1} 8 y d y = 4

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

= 2 π (16 - \frac{16}{3} - 4 + 2)

Simplify

2 π (16 - \frac{16}{3} - 4 + 2) : \frac{52 π}{3}

= \frac{52 π}{3}

\int \frac{e ^{2 x}}{e ^{x}} d x

(0, - 9), (7, - 8)

x \to 4 lim (x^{2} - 3 x + 2)

x \to 0 lim (6 x)

tan (e^{9 t}) + e^{t a n (9 t)}

What is the integral from 0 to 1 of 2pi(2-y)(8-8y^2) ?
The integral from 0 to 1 of 2pi(2-y)(8-8y^2) is (52pi)/3