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

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

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

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

\frac{648}{5}

Decimal

129.6

Solution steps

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

Expand

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

= \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

= \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}

= 243 - 162 + \frac{243}{5}

Simplify

243 - 162 + \frac{243}{5} : \frac{648}{5}

= \frac{648}{5}

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