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

Eliminate Absolutes

= \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}

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\int_{1}^{3} x^{2} - 4 d x

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

\int_{\frac{π}{6}}^{\frac{π}{3}} csc (x) d x

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