What is the integral from 0 to 3 of 9x^2-x^4 ?

The integral from 0 to 3 of 9x^2-x^4 is 162/5

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integral from 0 to 3 of 9x^2-x^4

\int_{0}^{3} 9 x^{2} - x^{4} d x

\frac{162}{5}

Decimal

32.4

Solution steps

\int_{0}^{3} 9 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} 9 x^{2} d x - \int_{0}^{3} x^{4} d x

\int_{0}^{3} 9 x^{2} d x = 81

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

= 81 - \frac{243}{5}

Simplify

81 - \frac{243}{5} : \frac{162}{5}

= \frac{162}{5}

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What is the integral from 0 to 3 of 9x^2-x^4 ?
The integral from 0 to 3 of 9x^2-x^4 is 162/5