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

The integral from 0 to 2 of (8r^2-2r^4) is 128/15

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

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

\frac{128}{15}

Solution steps

\int_{0}^{2} 8 r^{2} - 2 r^{4} d r

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} 8 r^{2} d r - \int_{0}^{2} 2 r^{4} d r

\int_{0}^{2} 8 r^{2} d r = \frac{64}{3}

\int_{0}^{2} 2 r^{4} d r = \frac{64}{5}

= \frac{64}{3} - \frac{64}{5}

Simplify

\frac{64}{3} - \frac{64}{5} : \frac{128}{15}

= \frac{128}{15}

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What is the integral from 0 to 2 of (8r^2-2r^4) ?
The integral from 0 to 2 of (8r^2-2r^4) is 128/15