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

The integral from 0 to 4 of 2pi(x)(4x-x^2) is (128pi)/3

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

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

\frac{128 π}{3}

Decimal

134.04128 \dots

Solution steps

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

Expand

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

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

\int_{0}^{4} 4 x^{2} d x = \frac{256}{3}

\int_{0}^{4} x^{3} d x = 64

= 2 π (\frac{256}{3} - 64)

Simplify

2 π (\frac{256}{3} - 64) : \frac{128 π}{3}

= \frac{128 π}{3}

What is the integral from 0 to 4 of 2pi(x)(4x-x^2) ?
The integral from 0 to 4 of 2pi(x)(4x-x^2) is (128pi)/3