What is the integral from 0 to 2 of 2pi(x+1)(12-6x) ?

The integral from 0 to 2 of 2pi(x+1)(12-6x) is 40pi

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integral from 0 to 2 of 2pi(x+1)(12-6x)

\int_{0}^{2} 2 π (x + 1) (12 - 6 x) d x

40 π

Decimal Notation

125.66370 \dots

Solution steps

\int_{0}^{2} 2 π (x + 1) (12 - 6 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π \cdot \int_{0}^{2} (x + 1) (12 - 6 x) d x

(x + 1) (12 - 6 x) = 6 (x + 1) (2 - x)

= 2 π \cdot \int_{0}^{2} 6 (x + 1) (2 - x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π 6 \cdot \int_{0}^{2} (x + 1) (2 - x) d x

Expand

(x + 1) (2 - x) : x - x^{2} + 2

= 2 π 6 \cdot \int_{0}^{2} x - x^{2} + 2 d x

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 6 (\int_{0}^{2} x d x - \int_{0}^{2} x^{2} d x + \int_{0}^{2} 2 d x)

\int_{0}^{2} x d x = 2

\int_{0}^{2} x^{2} d x = \frac{8}{3}

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

= 2 π 6 (2 - \frac{8}{3} + 4)

Simplify

2 π 6 (2 - \frac{8}{3} + 4) : 40 π

= 40 π

g (x) = 3^{(\frac{x}{2})}

x \to 2 lim (\frac{( 3 )}{x - 2})

\int sec^{7} (3 x) tan (3 x) d x

x \to 0 - lim (cot (x))

(x (csc (\frac{y}{x})) - y) d x + x d y = 0

What is the integral from 0 to 2 of 2pi(x+1)(12-6x) ?
The integral from 0 to 2 of 2pi(x+1)(12-6x) is 40pi