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

The integral from 0 to 4 of pi(4x-x^2) is pi 32/3

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

\int_{0}^{4} π (4 x - x^{2}) d x

π \frac{32}{3}

Decimal Notation

33.51032 \dots

Solution steps

\int_{0}^{4} π (4 x - x^{2}) d x

Take the constant out:

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

= π \cdot \int_{0}^{4} 4 x - 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

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

\int_{0}^{4} 4 x d x = 32

\int_{0}^{4} x^{2} d x = \frac{64}{3}

= π (32 - \frac{64}{3})

Join

32 - \frac{64}{3} : \frac{32}{3}

= π \frac{32}{3}

x sin (- 2 x) π

9 y^{^{''}} - 7 y = 0

\frac{d}{d x} (\frac{40}{x + 1})

\frac{( s + 2 )}{s ^{2} + 4}

What is the integral from 0 to 4 of pi(4x-x^2) ?
The integral from 0 to 4 of pi(4x-x^2) is pi 32/3