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

The integral from 0 to 2 of (4x^2-x^4) is 64/15

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

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

\frac{64}{15}

Decimal

4.26666 \dots

Solution steps

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

\int_{0}^{2} 4 x^{2} d x = \frac{32}{3}

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

= \frac{32}{3} - \frac{32}{5}

Simplify

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

= \frac{64}{15}

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What is the integral from 0 to 2 of (4x^2-x^4) ?
The integral from 0 to 2 of (4x^2-x^4) is 64/15