What is the integral from 0 to 3 of pi(9-x^2)^2 ?

The integral from 0 to 3 of pi(9-x^2)^2 is pi 648/5

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integral from 0 to 3 of pi(9-x^2)^2

\int_{0}^{3} π (9 - x^{2})^{2} d x

π \frac{648}{5}

Decimal

407.15040 \dots

Solution steps

\int_{0}^{3} π (9 - x^{2})^{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}^{3} (9 - x^{2})^{2} d x

Expand

(9 - x^{2})^{2} : 81 - 18 x^{2} + x^{4}

= π \cdot \int_{0}^{3} 81 - 18 x^{2} + x^{4} 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}^{3} 81 d x - \int_{0}^{3} 18 x^{2} d x + \int_{0}^{3} x^{4} d x)

\int_{0}^{3} 81 d x = 243

\int_{0}^{3} 18 x^{2} d x = 162

\int_{0}^{3} x^{4} d x = \frac{243}{5}

= π (243 - 162 + \frac{243}{5})

Simplify

π (243 - 162 + \frac{243}{5}) : π \frac{648}{5}

= π \frac{648}{5}

What is the integral from 0 to 3 of pi(9-x^2)^2 ?
The integral from 0 to 3 of pi(9-x^2)^2 is pi 648/5