What is the integral from 0 to 2 of 6y^2-2y^4+y^3 ?

The integral from 0 to 2 of 6y^2-2y^4+y^3 is 36/5

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integral from 0 to 2 of 6y^2-2y^4+y^3

\int_{0}^{2} 6 y^{2} - 2 y^{4} + y^{3} d y

\frac{36}{5}

Decimal

7.2

Solution steps

\int_{0}^{2} 6 y^{2} - 2 y^{4} + y^{3} d y

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} 6 y^{2} d y - \int_{0}^{2} 2 y^{4} d y + \int_{0}^{2} y^{3} d y

\int_{0}^{2} 6 y^{2} d y = 16

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

\int_{0}^{2} y^{3} d y = 4

= 16 - \frac{64}{5} + 4

Simplify

16 - \frac{64}{5} + 4 : \frac{36}{5}

= \frac{36}{5}

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What is the integral from 0 to 2 of 6y^2-2y^4+y^3 ?
The integral from 0 to 2 of 6y^2-2y^4+y^3 is 36/5