What is the integral from 0 to 6 of 2pi(y+2)(3-y/2) ?

The integral from 0 to 6 of 2pi(y+2)(3-y/2) is 72pi

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integral from 0 to 6 of 2pi(y+2)(3-y/2)

\int_{0}^{6} 2 π (y + 2) (3 - \frac{y}{2}) d y

72 π

Decimal

226.19467 \dots

Solution steps

\int_{0}^{6} 2 π (y + 2) (3 - \frac{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}^{6} (y + 2) (3 - \frac{y}{2}) d y

Simplify

(y + 2) (3 - \frac{y}{2}) : - \frac{y ^{2}}{2} + 2 y + 6

= 2 π \cdot \int_{0}^{6} - \frac{y ^{2}}{2} + 2 y + 6 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}^{6} \frac{y ^{2}}{2} d y + \int_{0}^{6} 2 y d y + \int_{0}^{6} 6 d y)

\int_{0}^{6} \frac{y ^{2}}{2} d y = 36

\int_{0}^{6} 2 y d y = 36

\int_{0}^{6} 6 d y = 36

= 2 π (- 36 + 36 + 36)

Simplify

= 72 π

x \to 0 - lim (csc (x) (x - 1))

\int_{- \infty}^{0} x^{2} e^{x} d x

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

\int_{1}^{3} e^{x} d x

\int_{3}^{6} \frac{x}{x ^{2} + 6 x + 13} d x

What is the integral from 0 to 6 of 2pi(y+2)(3-y/2) ?
The integral from 0 to 6 of 2pi(y+2)(3-y/2) is 72pi