What is the integral from 0 to 3 of 2pi(3-x)(9-x^2) ?

The integral from 0 to 3 of 2pi(3-x)(9-x^2) is (135pi)/2

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integral from 0 to 3 of 2pi(3-x)(9-x^2)

\int_{0}^{3} 2 π (3 - x) (9 - x^{2}) d x

\frac{135 π}{2}

Decimal

212.05750 \dots

Solution steps

\int_{0}^{3} 2 π (3 - x) (9 - x^{2}) 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}^{3} (3 - x) (9 - x^{2}) d x

Expand

(3 - x) (9 - x^{2}) : 27 - 3 x^{2} - 9 x + x^{3}

= 2 π \cdot \int_{0}^{3} 27 - 3 x^{2} - 9 x + x^{3} 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 (\int_{0}^{3} 27 d x - \int_{0}^{3} 3 x^{2} d x - \int_{0}^{3} 9 x d x + \int_{0}^{3} x^{3} d x)

\int_{0}^{3} 27 d x = 81

\int_{0}^{3} 3 x^{2} d x = 27

\int_{0}^{3} 9 x d x = \frac{81}{2}

\int_{0}^{3} x^{3} d x = \frac{81}{4}

= 2 π (81 - 27 - \frac{81}{2} + \frac{81}{4})

Simplify

2 π (81 - 27 - \frac{81}{2} + \frac{81}{4}) : \frac{135 π}{2}

= \frac{135 π}{2}

- 4 e^{- x} + 1

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f (x) = (1 + 2 x)^{2}, at x = 3

g

t^{2} + 5

What is the integral from 0 to 3 of 2pi(3-x)(9-x^2) ?
The integral from 0 to 3 of 2pi(3-x)(9-x^2) is (135pi)/2