What is the integral from 0 to 3 of 2pi(5-x)(9-x^2) ?

The integral from 0 to 3 of 2pi(5-x)(9-x^2) is (279pi)/2

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integral from 0 to 3 of 2pi(5-x)(9-x^2)

\int_{0}^{3} 2 π (5 - x) (9 - x^{2}) d x

\frac{279 π}{2}

Decimal Notation

438.25217 \dots

Solution steps

\int_{0}^{3} 2 π (5 - 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} (5 - x) (9 - x^{2}) d x

Expand

(5 - x) (9 - x^{2}) : 45 - 5 x^{2} - 9 x + x^{3}

= 2 π \cdot \int_{0}^{3} 45 - 5 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} 45 d x - \int_{0}^{3} 5 x^{2} d x - \int_{0}^{3} 9 x d x + \int_{0}^{3} x^{3} d x)

\int_{0}^{3} 45 d x = 135

\int_{0}^{3} 5 x^{2} d x = 45

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

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

= 2 π (135 - 45 - \frac{81}{2} + \frac{81}{4})

Simplify

2 π (135 - 45 - \frac{81}{2} + \frac{81}{4}) : \frac{279 π}{2}

= \frac{279 π}{2}

arcsin (x)

\frac{0.707}{s + 0.707}

\frac{d}{d x} (a \cdot cos (x) + b \cdot sin (x))

\int (2 (z + 1) e^{2 x - 2} + 2 y) d x

\frac{d}{d x} (A x e^{4 x})

What is the integral from 0 to 3 of 2pi(5-x)(9-x^2) ?
The integral from 0 to 3 of 2pi(5-x)(9-x^2) is (279pi)/2