What is the integral from 0 to 2 of 2pi(x+1)(4-2x) ?

The integral from 0 to 2 of 2pi(x+1)(4-2x) is (40pi)/3

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integral from 0 to 2 of 2pi(x+1)(4-2x)

\int_{0}^{2} 2 π (x + 1) (4 - 2 x) d x

\frac{40 π}{3}

Decimal Notation

41.88790 \dots

Solution steps

\int_{0}^{2} 2 π (x + 1) (4 - 2 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) (4 - 2 x) d x

(x + 1) (4 - 2 x) = 2 (x + 1) (2 - x)

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

Expand

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

= 2 π 2 \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 2 (\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 2 (2 - \frac{8}{3} + 4)

Simplify

2 π 2 (2 - \frac{8}{3} + 4) : \frac{40 π}{3}

= \frac{40 π}{3}

\int_{0}^{1} (\frac{3}{1 + t ^{2}}) d t

3 x cos (x)

- 4 - 9 x^{2}

2 \frac{d ^{2} y}{d x ^{2}} + 11 \frac{d y}{d x} - 6 y = 8 x

(x^{2} - 4) \frac{d y}{d x} + 4 y = (x + 2)^{2}

What is the integral from 0 to 2 of 2pi(x+1)(4-2x) ?
The integral from 0 to 2 of 2pi(x+1)(4-2x) is (40pi)/3