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

The integral from 0 to 4 of pi(x^2-4x)^2 is pi 512/15

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

\int_{0}^{4} π (x^{2} - 4 x)^{2} d x

π \frac{512}{15}

Decimal Notation

107.23302 \dots

Solution steps

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

Expand

(x^{2} - 4 x)^{2} : x^{4} - 8 x^{3} + 16 x^{2}

= π \cdot \int_{0}^{4} x^{4} - 8 x^{3} + 16 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} x^{4} d x - \int_{0}^{4} 8 x^{3} d x + \int_{0}^{4} 16 x^{2} d x)

\int_{0}^{4} x^{4} d x = \frac{1024}{5}

\int_{0}^{4} 8 x^{3} d x = 512

\int_{0}^{4} 16 x^{2} d x = \frac{1024}{3}

= π (\frac{1024}{5} - 512 + \frac{1024}{3})

Simplify

π (\frac{1024}{5} - 512 + \frac{1024}{3}) : π \frac{512}{15}

= π \frac{512}{15}

\int 5^{c o s (x)} sin (x) d x

\frac{d}{d y} (2 x^{3} + 6 x y^{2} - 3 x^{2} + 3 y^{2})

\frac{d y}{d x} = \frac{1 + 2 y}{1 + x}

1 + 2 t

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

What is the integral from 0 to 4 of pi(x^2-4x)^2 ?
The integral from 0 to 4 of pi(x^2-4x)^2 is pi 512/15