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

The integral from 0 to 2 of pi(x^2-4)^2 is pi 256/15

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

\int_{0}^{2} π (x^{2} - 4)^{2} d x

π \frac{256}{15}

Decimal Notation

53.61651 \dots

Solution steps

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

Expand

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

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

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

\int_{0}^{2} 8 x^{2} d x = \frac{64}{3}

\int_{0}^{2} 16 d x = 32

= π (\frac{32}{5} - \frac{64}{3} + 32)

Join

\frac{32}{5} - \frac{64}{3} + 32 : \frac{256}{15}

= π \frac{256}{15}

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What is the integral from 0 to 2 of pi(x^2-4)^2 ?
The integral from 0 to 2 of pi(x^2-4)^2 is pi 256/15