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

The integral from 0 to 2 of 2pix(8x(x-2)^2) is (256pi)/(15)

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integral from 0 to 2 of 2pix(8x(x-2)^2)

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

\frac{256 π}{15}

Decimal

53.61651 \dots

Solution steps

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

Take the constant out:

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

= 2 π 8 \cdot \int_{0}^{2} xx (x - 2)^{2} d x

Simplify

= 16 π \cdot \int_{0}^{2} xx (x - 2)^{2} d x

Expand

xx (x - 2)^{2} : x^{4} - 4 x^{3} + 4 x^{2}

= 16 π \cdot \int_{0}^{2} x^{4} - 4 x^{3} + 4 x^{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

= 16 π (\int_{0}^{2} x^{4} d x - \int_{0}^{2} 4 x^{3} d x + \int_{0}^{2} 4 x^{2} d x)

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

\int_{0}^{2} 4 x^{3} d x = 16

\int_{0}^{2} 4 x^{2} d x = \frac{32}{3}

= 16 π (\frac{32}{5} - 16 + \frac{32}{3})

Simplify

16 π (\frac{32}{5} - 16 + \frac{32}{3}) : \frac{256 π}{15}

= \frac{256 π}{15}

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

\int_{- 8}^{2} 16 - 6 x - x^{2} d x

\int_{0}^{\infty} ∣ sin (x) ∣ d x

\int_{- 5}^{5} - 25 - x^{2} d x

\int_{0}^{1} 7 x (x^{2} - 1)^{5} d x

What is the integral from 0 to 2 of 2pix(8x(x-2)^2) ?
The integral from 0 to 2 of 2pix(8x(x-2)^2) is (256pi)/(15)