What is the integral from 0 to 2 of (-x^4+2x^2+8) ?

The integral from 0 to 2 of (-x^4+2x^2+8) is 224/15

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integral from 0 to 2 of (-x^4+2x^2+8)

\int_{0}^{2} (- x^{4} + 2 x^{2} + 8) d x

\frac{224}{15}

Decimal

14.93333 \dots

Solution steps

\int_{0}^{2} - x^{4} + 2 x^{2} + 8 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} x^{4} d x + \int_{0}^{2} 2 x^{2} d x + \int_{0}^{2} 8 d x

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

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

\int_{0}^{2} 8 d x = 16

= - \frac{32}{5} + \frac{16}{3} + 16

Simplify

- \frac{32}{5} + \frac{16}{3} + 16 : \frac{224}{15}

= \frac{224}{15}

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What is the integral from 0 to 2 of (-x^4+2x^2+8) ?
The integral from 0 to 2 of (-x^4+2x^2+8) is 224/15