What is the integral from 0 to 2 of 2pi(9-x)(8-x^3) ?

The integral from 0 to 2 of 2pi(9-x)(8-x^3) is (984pi)/5

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integral from 0 to 2 of 2pi(9-x)(8-x^3)

\int_{0}^{2} 2 π (9 - x) (8 - x^{3}) d x

\frac{984 π}{5}

Decimal Notation

618.26543 \dots

Solution steps

\int_{0}^{2} 2 π (9 - x) (8 - x^{3}) 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} (9 - x) (8 - x^{3}) d x

Expand

(9 - x) (8 - x^{3}) : 72 - 9 x^{3} - 8 x + x^{4}

= 2 π \cdot \int_{0}^{2} 72 - 9 x^{3} - 8 x + x^{4} 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}^{2} 72 d x - \int_{0}^{2} 9 x^{3} d x - \int_{0}^{2} 8 x d x + \int_{0}^{2} x^{4} d x)

\int_{0}^{2} 72 d x = 144

\int_{0}^{2} 9 x^{3} d x = 36

\int_{0}^{2} 8 x d x = 16

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= 2 π (144 - 36 - 16 + \frac{32}{5})

Simplify

2 π (144 - 36 - 16 + \frac{32}{5}) : \frac{984 π}{5}

= \frac{984 π}{5}

\int_{120}^{150} 0.1281 \cdot x \cdot e^{- \frac{x}{100}} d x

4 y + y^{^{''}} = 0

(\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 9 x^{3})

\int \frac{2}{( 1 + x ^{2} ) ^{2}} d x

\frac{d ^{2} y}{d t ^{2}} + 10 \frac{d y}{d t} + 25 y = 0

What is the integral from 0 to 2 of 2pi(9-x)(8-x^3) ?
The integral from 0 to 2 of 2pi(9-x)(8-x^3) is (984pi)/5